chapter 2 vibrational spectroscopy
TRANSCRIPT
Chapter 2
Vibrational Spectroscopy
2.1 Introduction
2.1.1 Infrared (IR) Spectroscopy
The absorption of infrared (IR) radiation causes excitation of vibrations of the
atoms of a molecule or the crystal lattice and causes bands in the spectra which are
generally presented in the unit wave number ~n in cm�1 (wavelength l was used in
the older literature). Commonly, the symbol n is used in vibrational spectroscopy
instead of the symbol ~n in the energy scale and they are sometimes named
vibrational frequencies which are measured in the maxima of absorption bands
whereas this is correct only for small oscillation strength.
The infrared range of the electromagnetic spectrum is divided into three regions,
named after their relation to the visible spectrum:
– Near-infrared (NIR) (wave number ranges from 14,000 to 4,000 cm�1 and
wavelength ranges from 0.8 to 2.5 mm) lying adjacent to the visible region
excites so-called overtone or higher harmonic vibrations (higher harmonics).– Mid-infrared (wave number ranges from approximately 4,000 to 400 cm�1
and wavelength ranges from 2.5 to 25 mm) excites mainly fundamental
vibrations. This part of the infrared range may be used to study the structure of
molecules that we shall be concerned with in this book. In general, the name
“IR-spectroscopy” conventionally refers to the mid-IR region.
– Far-infrared (wave number ranges from approximately 400 to 10 cm�1 and
wavelength ranges from 25 to 1,000 mm) is the lowest-energy region which
excites mainly lattice vibrations or can be used for rotational spectroscopy.
A molecule can vibrate in many ways, and each way is called a vibrationalmode. Let us start with a simple diatomic molecule A–B, such as CO or HCl. The
wave number of absorbance n can be calculated by Eq. 2.1 derived by the model of
the harmonic oscillator
M. Reichenb€acher and J. Popp, Challenges in Molecular Structure Determination,DOI 10.1007/978-3-642-24390-5_2, # Springer-Verlag Berlin Heidelberg 2012
63
n ¼ 1
2 p c
ffiffiffif
m
s(2.1)
in which c is the light velocity, f is the force constant in the atomic scale and the
spring constant for the macroscopic model, respectively, and m is the reduced mass
defined by Eq. 2.2
m ¼ mA � mB
mA þ mB: (2.2)
Note that the force constant is a criterion for the strength of the chemical bond in
the molecule A�B. Therefore, the stronger the chemical bond (electronic effect)
and the smaller the reduced mass m (mass effect), the higher the wave number of the
absorption band n.The relation in Eq. 2.1 can be applied not only for the simple diatomic molecule
but also for some structural moieties in more complex molecules as long as
coupling between the atoms can be neglected, in other words, at so-called charac-teristic vibrations which are explained in Sect. 2.3.2. Such vibrations are provided,
for example, by multiple bonds in the neighborhood of single bonds.
Electronic effects on the position of the absorption band
The force constant increases with the bonding order, for example, fC-X < fC¼X< f
C�X. Therefore, the vibrational frequencies increase for vibrations of equal atoms
X according to Eq. 2.1:
nðC�CÞ � 1;000 cm�1 < nðC¼CÞ � 1; 650 cm�1 < nðC�CÞ � 2; 250 cm�1
nðC�OÞ � 1; 100 cm�1 < nðC¼OÞ � 1; 700 cm�1:
Note that Eq. 2.1 enables the recognition of the real strength of a chemical bond.
For example, the hydrocarbon bond C�H is famously a single bond, but this is onlya generalization. The neighborhood of oxygen at the carbon atom diminishes the
electron density and, hence, the force constant and bonding strength, resulting in a
smaller vibrational frequency according to Eq. 2.1. This is true for the following
structural moieties: n(AlkylC�H) � 3,000 cm�1 and n(O¼C�H) � 2,750 cm�1.Thus, the C�H-bond is slightly stronger in the alkyl group.
Let us now consider I- and M-effects on the vibrational frequencies.
The vibrational frequencies of the C¼O double bond, n(C¼O) of compounds of
the type CH3C(C¼O)�X are: 1,742 cm�1 (for X ¼ H), 1,750 cm�1 (for X ¼ Cl),
and 1,685 cm�1 (for X ¼ phenyl). Thus, there are differences concerning the
strength of the C¼O bond due to the nature of the chemical bond: The –I-effect
of the Cl atom diminishes the electron density at the C atom which is equalized by
contribution of electron density from the free electron pair at the neighboring O
atom. Thus, the bond order for the C¼O bond is increased as is shown by the
resonance structures C�C(¼O)Cl $ C�C�O+Cl� which increases the force
64 2 Vibrational Spectroscopy
constant and, hence, a higher vibrational frequency is observed in carboxylic acid
chlorides.
Otherwise, a substituent with a + M-effect (phenyl) causes conjugation of the
C¼O group as Structure 2.1 shows
O
–
O+
2.1
resulting in a smaller force constant and, hence, a smaller vibrational frequency.
Conjugation increases the thermodynamic stability of the molecule butdecreases the bond strength and, hence, the force constant which results in smallervibrational frequencies.
Conversely, the vibrational frequency obtained experimentally by the infrared
spectrum provides hints to the strength of the chemical bond.
Note that coupling of various vibrations results in absorption bands of higher and
lower frequency in the spectrum which are not caused by electronic effects. The
transitions 3 and 4 caused by coupling of transitions 1 and 2 are shown in a simple
energy diagram in Fig. 2.1.
Mass effects on the position of the absorption band
The smaller the mass of the atoms the higher the vibrational frequency, which is
demonstrated by the C–H and C–D stretching frequencies,
nðC�HÞ � 3;000 cm�1; nðC� DÞ � 2; 120 cm�1:
1
3 4
2
Energy
Fig. 2.1 Schematic energy diagram for the coupling of transitions 1 and 2 resulting in the
transitions 3 and 4 with corresponding absorption bands in the infrared spectrum
2.1 Introduction 65
The vibrational frequencies of heavier isotopes can markedly differ and they can
occur with proper intensities for atoms with abundant isotopes.
As mentioned above, Eq. 2.1 is approximately valid for characteristic vibrations
in multiatomic molecules. For such a structural moiety AB, for example, the C¼Ogroup of a carbonyl compound, the force constant f can be calculated according to
the so-called “two-mass model” by Eq. 2.3
f in N�1 ¼ 5:891 � 10�5 � n2
MA�1 þMB
�1 (2.3)
in which n is the wave number of the corresponding experimentally observed
absorption band in cm�1, and MA and MB are the relative atomic masses of the
atoms A and B, respectively.
Challenge 2.1
Calculate the force constant of the P¼O group in OPCl3. The strong infrared
absorption band of the P¼O stretching vibration is observed at 1,290 cm�1.
Solution to Challenge 2.1
The force constant is f(P¼O) ¼ 1,035 N m�1 calculated by Eq. 2.1 with
n ¼ 1,290 cm�1, MP ¼ 31, and MO ¼ 16.
In the same manner the force constants can be calculated for carbon�carbonbonds and bonds in inorganic species resulting in the following rules:
– The relation of the force constants of the single, double, and triple bonds f(C�C)� 500 N m�1, f(C¼C) � 1,000 N m�1, f(C�C) � 1,500 N m�1 is 1:2:3.
– The force constant increases with increasing effective nuclear charge, for exam-
ple, fS-O(SO32�) ¼ 552 N m�1 < fS-O(SO4
2�) ¼ 715 N m�1.– The force constant increases with increasing s-character of the hybrid orbital:
fC-H(sp3C�H) < fC-H(sp
2C�H) < fC-H(spC�H).It is well known, however, that not all aspects of the atomic level can be
described by the macroscopic ball-and-spring model. Instead the quantum mechan-
ical formalism must be applied resulting in discrete energy levels
En ¼ h � n � vþ 1
2
� �(2.4)
in which v is the vibrational quantum number. According to this selection rule, only
transitions are allowed with Dv ¼ �1. Because for most vibrations of organic
66 2 Vibrational Spectroscopy
molecules nearly exclusively the ground state v ¼ 0 is occupied at ambient
temperatures, the experimentally observed infrared absorption band of a two-
atomic molecule is caused by the transition v ¼ 0 ! v ¼ 1 which is named the
fundamental vibration.Furthermore, the harmonic oscillator should be substituted by the anharmonic
oscillator model. This substitution leads to a shape of the potential curve which is
asymmetric and the right part of which converges to the dissociation energy.
Therefore, the energy levels are no longer equidistant but decrease with increasing
vibrational quantum number v. Moreover, the selection rule for the harmonic
oscillator is no longer strongly valid; transitions with Dv ¼ �2, �3 . . .. canadditionally occur, albeit with lower intensity, which are named overtones or higherharmonics. Thus, overtones are, in a first approximation, the whole number
multiples of the fundamentals. Because of the anharmonicity of vibration the first
overtone is somewhat smaller in frequency than the double value of the
corresponding fundamental and so on.
The intensity of an absorption band is determined by the strength of the electro-
magnetic absorption process which is mainly caused by the change of the electric
dipole moment for a vibrational transition induced by infrared radiation. Note that
the probability of a transition with Dv ¼ �2 is approximately tenfold smaller than
that with Dv ¼ �1. The absorption coefficients a of fundamentals are in the range
1–100 L mol�1 cm�1, therefore, overtones appear with low intensities or are
missing in a routine spectrum.
In molecules with more than two atoms anharmonicity can additionally lead to
combination modes which refer to the simultaneous excitation of two vibrational
modes at one frequency which can lead to additional peaks and further complicate
infrared spectra. Some of them are meaningful for structural analysis, for example,
for the determination of aromatic substituent types (see Sect. 2.3).
Challenge 2.2
For solutions to Challenge 2.2, see extras.springer.com.
1. The experimentally observed N¼O fundamentals n(N¼O) of nitrosyl
compounds O¼NX are: 1,800 cm�1 (for X ¼ Cl, gaseous state),
1,948 cm�1 (for X ¼ Cl, solid state), and 2,387 cm�1 (for X ¼ BF4�).
Characterize the chemical bonding in these compounds and formulate the
structure.
2. Calculate the force constants f(C¼X) according to the two-mass model for
the C¼X bond for X ¼ C and X ¼ O, on the basis of the experimentally
observed fundamentals: H2C¼O: n(C¼O) ¼ 1,742 cm�1, H2C¼CH2:
n(C¼C) ¼ 1,645 cm�1.3. Calculate the force constant by means of the simple two-mass model for the
C�H and C�C bonds in H�C�N with the experimentally observed
fundamentals n(C�H) ¼ 3,311 cm�1 and n(C�C) ¼ 2,097 cm�1,respectively. The exact values obtained by a normal mode analysis are
(continued)
2.1 Introduction 67
f(C�H) ¼ 582 N m�1 and f(C�C) ¼ 1,785 N m�1. What is the percentage
deviation between the exact and the approximate values? What knowledge
can be gained from this result?
4. The stretching frequency of the 16O-isotope is experimentally determined
to be n(16O�16O) ¼ 832 cm�1. At which wave number can the stretching
frequency for the 18O-isomer, n(18O�18O) be expected?
2.1.2 Raman Spectroscopy and Some Applications
Raman spectra result from inelastic scattering of monochromatic light, usually from
a laser in the visible, near-infrared, and near-ultraviolet range. Such lasers are for
example: Ar ion with wavelengths at 514, 488, and 458 nm, He/Ne with
wavelengths at 628, 578, and 442 nm, or Nd/YAG with wavelengths at 523 and
1,052 nm. Note that the last wavelength lies in the NIR range.
The Raman effect occurs when monochromatic light impinges upon a molecule
and interacts with the electron cloud and the bonds of this molecule. For the
spontaneous Raman effect to take place the molecule is excited by a photon from
the ground state to a virtual energy state. The molecule can relax and return to the
ground state by emission of a photon whose energy is the same as that of the
exciting radiation resulting in elastic Rayleigh scattering. But a very small part of
the scattered light can also have frequencies that are smaller than those of the
elastically scattered part because a part of the energy of the incoming photons was
employed to excite molecules to a higher vibrational state v ¼ 1. As a result, the
emitted photons will be shifted to lower energy, i.e., light with a lower frequency.
This shift is designated the Stokes shift. The difference in energy between the
original state and the final state corresponds to a vibrational mode far from the
excitation wavelength.
If the process starts from a vibrationally excited state v ¼ 1 and relaxes to the
ground state v ¼ 0, then the emitted photons will be shifted to higher frequency
which is designated the Anti-Stokes shift. Compared with the Stokes-shifted light
the Anti-Stokes shifted radiation has a lower intensity because of the small popula-
tion of the vibrationally excited state compared to that of the ground state of the
molecule at ambient temperatures. These processes are illustrated in a simple
energy level diagram in Fig. 2.2. As Fig. 2.2 shows, the infrared absorption yields
similar, but often complementary information. However, whereas the vibrational
frequency directly corresponds to the absorption band in the infrared spectrum the
vibrational frequency in the Raman spectra corresponds to the difference betweenRayleigh and Stokes lines in the Raman spectrum. Note that the Raman effect is a
scattering phenomenon and may not be confused with fluorescence whose emission
starts from a discrete and not a virtual energy level (see Sect. 3.7).
Besides spontaneous Raman spectroscopy there are a number of advanced types
of Raman spectroscopy, for example, surface-enhanced Raman (SERS), resonance
68 2 Vibrational Spectroscopy
Raman spectroscopy (RRS), coherent anti-Stokes Raman spectroscopy (CARS),
and others which are far beyond the scope of this book.
In principle, the registration of a Raman spectrum is simple: The sample is
illuminated with an appropriate laser beam. The stray light is collected by a lens and
sent through a monochromator. The elastic Rayleigh scattering light is filtered out
while the rest of the collected light is dispersed onto a detector.
Although, in general, the organic chemist will prefer infrared spectroscopy for
the registration of vibrational spectra, spontaneous Raman spectroscopy has a
number of advantages; some of them will be given in the following text.
• The frequency of the excitation light can be chosen arbitrarily (within the limits
of available laser frequencies). Thus, glass cuvettes can be applied for excitation
with visible light instead of the expensive alkali discs used in infrared
spectroscopy.
• Whereas the excitation of vibrations by infrared radiation is connected with
changing the electrical dipole moment, in Raman spectroscopy only such
vibrations are active that change the induced dipole moment (polarizability).Thus, polar bonds provide intense absorption bands whereas intense Raman
lines are commonly observed from nonpolar bonds, such as C�C, C¼C,C�C, S�S, N¼N, and others.
• Infrared and Raman spectroscopy are complementary for the investigation of
symmetrical species, mostly present in inorganic chemistry, for the determina-
tion of allmodes whereas there are also modes that cannot be excited either in IR
or in Raman spectroscopy. Note that, according to the selection rules explained
V = 0
V = 1
V = 2
V = 3
V ′ = 0
V ′ =10
Energy
I
Ground state
II
Virtual energy state
IV
Infrared absorption
III
Fig. 2.2 Energy level diagram showing Rayleigh (I), Stokes (Raman) (II), and anti-Stokes
(Raman) (III) scattering as well as IR-absorption (IV)
2.1 Introduction 69
in Sect. 2.1.3, symmetrical modes can only be observed in the Raman spectrum
if the molecule possesses an inversion center.
• Because of the small intensities of the H–O stretching modes of water and
alcohols, Raman spectroscopy is excellently suitable for alcoholic or aqueous
solutions which is important, for example, in biological systems, such as the
determination of the presence of S–S bridges in proteins, the symmetrical
O�P�O mode of nuclear acids, or for the in vivo denaturation of proteins by
means of Raman microscopy.
• C�C modes in rings can be better recognized and assigned because of their
higher intensity in comparison to the infrared absorption bands.
• Raman spectra mostly show simpler patterns because of often-missing overtones
and combination modes.
• Note that the Raman spectrum presents the whole range of molecular and lattice
modes including those at low frequency n < 200 cm�1 for which an extra setup
(far-infrared spectrometer with PE cuvette) may be necessary in infrared
spectroscopy.
Examples for applications in the far-infrared range are investigations of:
– The low-frequency vibrations of metal–metal modes, such as, for example
n(Hg�Hg) ¼ 169 cm�1 in Hg2(NO3)2, and n(Mn�Mn) ¼ 157 cm�1 in
Mn2(CO)10;
– Lattice modes in solids in order to characterize or to distinguish polymorphic
forms;
– Stretching (n) and deformation (d) modes of heavy atoms, for example,
[CdI4]2�: 145 (nas), 117 (ns), 44 (d1), 23 (d2).
• The polarization of Raman scattered light also contains important information.
This property can be measured using plane-polarized laser excitation and a
polarization analyzer as shown in Fig. 2.3.
z-polarized excitation light induces a dipole moment in the molecule. Scattered
light travelling in the x-direction is also z-polarized light I|| (a).
z
y
x I||
a
P
A
z
x
I⊥
b
y
Fig. 2.3 Arrangement for measuring the polarization ratio of Raman scattered light. P polarizer,
A analyzer, I||, I⊥ intensity of Raman scattered light with the analyzer set parallel and perpendicular
to the excitation plane, respectively
70 2 Vibrational Spectroscopy
The polarization plane of the excitation laser light as well as the direction of the
induced dipole is rotated around 90� in (b). Thus, in case of fully symmetric modes
of optically isotropic substances, no z-polarized light is emitted travelling in the
x-direction which means, the intensity I⊥ is approximately zero.
Spectra acquired with an analyzer set both parallel (a) and perpendicular (b) to
the excitation plane (P) can be used to calculate the polarization ratio, which is
appropriate to experimentally recognize modes that are totally symmetrical (see
Sect. 2.1.3). The degrees of polarization r are calculated by the ratio of the
intensities measured employing the perpendicular and parallel analyzer set
according to Eq. 2.5
r ¼ I?Ijj
: (2.5)
Totally symmetrical modes provide polarized (p) Raman lines for which 0< r< 34
is valid. All other modes are depolarized (dp) with r � 34.
Figure 2.4 shows the Raman spectra of carbon tetrachloride measured with an
analyzer set both parallel (I||) and perpendicular (I⊥) to the excitation plane and the
polarization state is denoted by p and dp, respectively. As the degrees of polariza-tion suggest, the Raman line at approximately 470 cm�1 is polarized. Thus, it must
be assigned to the symmetric C–Cl stretching vibration. The connection between
symmetry of modes, Raman activity, and the Raman spectra is the subject of the
next section.
In addition to polarized IR spectroscopy, polarized Raman spectroscopy also can
be used for example, in solid-state physics to determine the orientation of the
crystallographic orientation of an anisotropic crystal and to understand macromo-
lecular orientations in crystal lattices and liquid crystals. In most cases, however,
Irel
100300500700
n in cm-1
dp
dp dp
p
I⊥
I||
Fig. 2.4 The polarized Raman spectra of carbon tetrachloride measured with an analyzer set both
parallel (I||) and perpendicular (I⊥) to the excitation plane. p polarized Raman lines, dp depolarizedRaman lines
2.1 Introduction 71
the application is more complex than in IR spectroscopy. Both methods have their
merits in polymorph analysis.
2.1.3 Symmetry, Selection Rules, and Applications
A nonlinear molecule with N atoms has 3 � N – 6 normal modes also called degrees
of freedom in contrast to when the molecule is linear. Then only 3 � N – 5 normal
modes are possible. As an example, the nonlinear molecule BF3 has 3 � 4 – 6 ¼ 6
degrees of freedom or normal modes while the linear CO2 has 3 � 3–5 ¼ 4 degrees
of freedom.
There are two principal types of molecular vibrations: stretching modes and
bending modes. A stretching vibration is characterized by movement along the
bond axis with increasing and decreasing of the interatomic distance. Stretching
vibrations are designated by the symbol n. A bending vibration consists of a changeof the bond angle between bonds or the movement of a group of atoms with respect
to the remainder of the molecule with an accompanying change of bond angle.
A movement in-plane is designated by the symbol d and g is used for out-of-planemovements (doop vibrations). Note that normal mode vibrations may not change the
center of gravity in the molecule.
The number of possible stretching vibrations is equal to the number of bonds andthe number of bending vibrations is the difference of the total 3 � N – 6(5) and thenumber of stretching modes.
Note that not only the “nonoscillating” molecule but also all modes of the
molecule have a symmetry which means that they can be assigned to one of the
symmetry classes of the point group to which every molecule belongs (for details,
see the specialized literature on group theory). The algorithm for the determination
of the point group of any molecule is presented in Table 6.7 and the symmetry
classes of the normal modes are listed in Table 6.9 for the most important symmetry
groups.
The principle of the determination of the symmetry properties of normal modes
will be described by the example of a planar compound AB3 to which belongs, for
example, the molecule BF3 or the ion NO3�.
According to the procedure given in Table 6.7 the triangular planar species AB3
belongs to the point group D3h. In order to determine the symmetry classes of the
fundamental modes, expediently, the base set of the so-called inner coordinates areused. In distinction from the Cartesian coordinates which include the total 3 � Nmodes, inner coordinates only include the possible vibrations of the molecule. The
inner coordinates of the stretching vibrations should be denoted by Dri, those of thebending in-plane by Dai, and the bending out-of-plane by Dgi. The coordinates are
defined in such a way that all possible movements of the molecule are included with
the consequence that the bending movements are over-determined. Three angles
cannot be simultaneously altered; the alteration of two angles fixes the third angle.
Thus, more representations are obtained as alterations of angles are possible.
72 2 Vibrational Spectroscopy
Therefore, without justification, a totally symmetrical representation G1 must be
eliminated.
Next, only the character of the transformational matrix has to be determined for
the base set of the inner coordinates. The following rule should be applied:
Only that coordinate of the base set provides a contribution to the character of the
transformational matrix that does not change its position at the execution of the respective
symmetry operation.
By applying this procedure, one obtains a reducible base set of the transforma-
tion matrix which can be decomposed into the set of irreducible representations Ziby means of the reduction formula (Eq. 2.6)
Zi ¼ 1 h= �X
gk � wrðKÞ � wirrðKÞ (2.6)
in which h is the order of the group (which is equal to the number of all symmetry
operations), gk is the number of the elements in each class K, and wr(K) and wirr(K)are the number of reducible and irreducible representations of the class K, respec-tively. The symmetry operations are listed in the character table which is given in
Table 6.8 for some symmetry groups.
Let us now continue the problem for AB3 given above. The molecule AB3 has
six normal modes: As the molecule has three bonds three stretching vibrations
result. Therefore, three normal modes remain belonging to the bending modes
which can be divided into in-plane (d) and out-of-plane (g) bending modes for a
planar molecule. The base set of the inner coordinates divided as described are
presented in Fig. 2.5.
The characters of the transformational matrix assigned to the three types of
modes according to Fig. 2.5 are listed in Table 2.1. They are obtained by applying
the rules given above.
A
B
BB
I II III
A
Δg2
A
B
BB
Δg1Δa2
Δa1
Δa3Δr3
Δr2Δr1
Fig. 2.5 Base set for the stretching (I), bending in-plane (II), and bending out-of-plane vibrations,
for the triangular planar molecule AB3 of the point group D3h
Table 2.1 Characters of the transformational matrix assigned to the three types of modes
(Fig. 2.5) for a triangular plane molecule AB3 belonging to the point group D3h
D3h E 2 C3 3 C02 sh 2 S3 sv Symmetry classes
w(Dri) 3 0 1 3 0 1 ) A1
0+ E
0
w(Dai) 3 0 1 3 0 1 ) A1
0+ E
0(�A1
0)
w(Dgi) 2 2 0 0 0 2 ) A1
0+ A2
00(�A1
0)
2.1 Introduction 73
The last column contains the irreducible representations calculated by Eq. 2.6
with the reducible ones given in columns 2–7. A totally symmetrical class A1
0must
be eliminated for the bending modes as explained above.
As Table 2.1 shows, the following six vibrations are obtained for a triangular
planar molecule AB3 with the symmetry D3h. (Note that the symbol G is commonly
used for the representation of a class).
Stretching vibrations : Gn ¼ A10 þ E0 (2.7)
Bending vibrations in - plane : Gd ¼ E0 (2.8)
Bending vibrations out - of - plane : Gg ¼ A200: (2.9)
Although six vibrations are theoretically possible only four vibrations can be
experimentally observed. The reason is that modes of the symmetry class E are
twofold degenerated which means that both of the degenerated modes are observed
as a single infrared absorption band or Raman line. Modes belonging to the
symmetry class E consist of two energetic modes perpendicular to each other.
Considering this fact, there are three stretching (Eq. 2.7), two bending in-plane
(Eq. 2.8), and one bending out-of-plane (Eq. 2.9) vibrations, in summary six
vibrations as calculated by 3 � 4–6.The fundamental modes can be illustrated by so-called distortion vectors as is
shown in Fig. 2.6 for the D3h molecule AB3. Note that the sum of these vectors may
not change the center of gravity of the molecule as this would be caused by a
translational motion.
Bending vibration in-plane
Symmetrical stretching vibration, ns Asymmetrical stretching vibration, nas
Bending vibration out-of-plane
A A A
A A A+
– –
–
ns (A1′) nas (E′)
g (A2″)d(E′)
Fig. 2.6 Diagram of the fundamental modes of a molecule AB3 with D3h symmetry
74 2 Vibrational Spectroscopy
The symmetry properties of molecules/ions of other symmetry groups can be
obtained by the same procedure. The results for the most important structures are
summarized in Table 6.9. Thus, the symmetry properties of the fundamental modes
of a pyramidal molecule AB3, for example NH3, with the symmetry group C3v are:
Stretching vibrations : Gn ¼ A1 þ E (2.10)
Bending vibrations : Gd ¼ A1 þ E: (2.11)
The question at hand is the following: Can one experimentally decide which of
the two possible structures of the molecule AB3 is present by means of the infrared
and Raman spectroscopy: a planar structure with D3h or pyramidal structure with
C3v symmetry? In order to answer this question the selection rules of vibrationalspectroscopy must be employed.
2.1.3.1 Selection Rules in Vibrational Spectroscopy
A vibration is only infrared active, i.e., it can only be observed as an infrared
absorption band if the corresponding vibration changes the electric dipole moment
m (note that a dipole moment need not exist in the equilibrium structure, but must
then be built up by the vibration). For such a transition the intensity is determined
by Einstein´s transition moment ~M which is proportional to the transition moment
integral in the volume dt
~M �ðC0~mCe dt (2.12)
in which C0 and Ce are the wave functions of the ground and the excited state,
respectively and~m is the dipole moment operator. The transition moment integral in
Eq. 2.12 does not equal zero and hence, the transition has a certain probability if the
integrant of the transition moment integral belongs to or contains the totally
symmetric representation G1.
In vibrational spectroscopy transitions are observed between different vibra-
tional states. In a fundamental vibration the molecule is excited from its ground
state (v ¼ 0) to the first excited state (v ¼ 1). Because the ground state belongs to
the totally symmetrical representation the integrant of the transition moment inte-
gral contains only the totally symmetrical representation G1 if the symmetry of the
excited state wave function is the same as the symmetry of the transition moment
operator. This fact gives rise to the selection rule in infrared spectroscopy.
2.1 Introduction 75
2.1.3.2 Selection Rule for Infrared Spectroscopy
A vibration can be observed in the infrared spectrum, i.e., a vibration is infraredactive, if the vibration is accompanied by a change in the electric dipole moment. Thisis the case if the symmetry of the respective vibration belongs to the same symmetryclass as one of the dipolemoment operators. Vibrations are infrared inactive if there isno change of the dipole moment with changing bond distance or angle.
The symmetry classes of the dipole moment operators are listed by x, y, and z, inthe second to last column of the character table (Table 6.8) for the point group
belonging to the molecule.
2.1.3.3 Selection Rule for Raman Spectroscopy and Alternative
Forbiddance
The Raman effect requires a changing induced dipole moment, i.e., the electric
polarizability which is the relative tendency of a charge distribution, like the
electron cloud of an atom or molecule, to be distorted from its normal shape by
an external electric field, while the molecule undergoes a vibration. The symmetry
of the induced dipole moment components are listed as the binary products in the
last column of the character table.
Raman active vibrations, i.e., vibrations which can be observed as Raman linesin the spectrum can only be such modes which are accompanied by a change of theinduced dipole moment or the polarizability. Such vibrations belong to the samesymmetry classes as the polarizability tensors listed in the last column of thecharacter table as binary products.
Alternative Forbiddance
Vibrations of molecules or ions having an inversion center i are either infrared(asymmetric modes) or Raman active (symmetric modes).
2.1.3.4 Application of the Selection Rules
After having discussed the selection rules we can come back to the question if one
can distinguish between the D3h and C3v structures for an AB3 molecule using
vibrational spectroscopy.
Table 2.2 shows the results of the application of the selection rules for both
structures of AB3-type molecules with D3h and C3v symmetry. The sign (+) means,
that the vibration is allowed and (�) means that the vibration is forbidden.
According to Table 2.2, all six fundamental modes are active both in the IR- and
the Raman spectrum for a molecule with pyramidal structure (C3v symmetry). But
76 2 Vibrational Spectroscopy
only three bands and lines are observed in the infrared and Raman spectra, respec-
tively for the planar structure D3h where only two vibrations coincide. Thus, the
application of vibrational spectroscopy can help to unambiguously distinguish
between D3h and C3v symmetry of an AB3 molecule or ion.
Note that sometimes the numbers of active vibrations are equal in the infrared
and Raman spectra. In this case, the numbers of polarized Raman lines can be
helpful in order to differentiate between possible structures. Bear in mind, only
totally symmetrical vibrations of the symmetry classes A1 or A01 provide polarized
Raman lines.
Challenge 2.3
For solutions to Challenge 2.3, see extras.springer.com
Determine the symmetry properties of the fundamental vibrations of
molecules BAX2 (for example, O¼CCl2 or O¼SCl2) with a planar and an
angled structure. Apply the selection rules and answer whether both
structures can be distinguished by means of vibrational spectroscopy.
2.1.4 Fermi Resonance
Most bands in the infrared spectrum arise from fundamental modes v ¼ 0! v ¼ 1,
from overtones v ¼ 0 ! v ¼ 2, or from combination modes which involve more
than one normal mode. There exist, however, absorption bands caused by a special
combination, namely the Fermi resonance. The phenomenon of Fermi resonance
can arise when two modes are similar in energy resulting in an unexpected shift in
energy and intensity of absorption bands.
Two conditions must be satisfied for Fermi resonance:
– The two vibrational states of a molecule transform according to the same
irreducible representation of the point group of the molecule.
– The transitions have accidentally almost the same energy.
Fermi resonance causes the high-energy mode to shift to higher energy and the
low-energy mode to shift to even lower energy. Furthermore, the weaker mode
gains intensity and the intensity of the more intense band decreases accordingly.
Table 2.2 Application of the selection rules to the vibrations of a molecule AB3 having a
triangular plane (D3h) and a pyramidal structure (C3v). The signs mean: (+): The transition is
allowed and the vibration is observed, (–): the transition is forbidden and the vibration cannot be
observed, and (p): the Raman line is polarized
D3h IR Raman C3v IR Raman
ns(A10) � + (p) ns(A1) + + (p)
nas(E0) + + nas(E) + +
d(E0) + + ds(A1) + + (p)
g(A2
00) + � das(E) + +
2.1 Introduction 77
Thus, Fermi resonance results in a linear combination of parent modes and does not
lead to additional bands in the spectrum.
Infrared absorption bands caused by Fermi resonance can be valuable for struc-
tural analysis. For example, two intense bands between 2,850 and 2,700 cm�1
are typical for the formyl group O¼C–H caused by Fermi resonance of the overtone
of the weak H–C¼O bending mode at approximately 1,390 cm�1 with the H–C
stretching mode. The occurrence of two (sometimes only one) bands in this region
unambiguously indicates the presence of a formyl group in the molecule.
A further example is the double infrared absorption band of the C¼O stretching
vibration of aryl acid chlorides, Aryl–C(¼O)Cl caused by coupling of the C¼Ostretching vibration with the overtone of the Cl–C¼O bending modes, d(Cl–C¼O)which lies in the same range. Thus, aryl acid chlorides can be distinguished from
alkyl acid chlorides by Fermi resonance. Further examples for Fermi resonances
will be found in Sect. 2.3.
2.2 Interpretation of Vibrational Spectra of Small Molecules
and Ions
The objective of the interpretation of vibrational spectra of small molecules and
ions differs from that of organic molecules and can be mainly divided into three
problems.
2.2.1 Determination of the Structure (Symmetry Group)of Molecules and Ions
The vibrational spectroscopic determination of the structure of a molecule or ion is
realized by the following steps:
1. Determination of the symmetry (or point) group of a postulated structure by
means of the algorithm given in Table 6.7.
2. Application of the selection rules on the fundamental modes the symmetries of
which are taken from Table 6.9.
3. Assignment of the experimentally observed infrared absorption bands and
Raman lines by means of the following rules:
(a) The wave numbers of stretching vibrations are always higher than those of
bending vibrations because the rhythmical movement along the bond axis
requires higher energy than changing of the bond angle.
(b) Consideration of the force constants and bond order, as well as the masses
for the assignment of the stretching modes according to the relation in
Eq. 2.1.
78 2 Vibrational Spectroscopy
(c) Consideration of the selection rules and the degrees of polarization of the
Raman lines.
(d) Comparison of the vibrational spectra of similar molecules or ions
concerning structure, force constant, and masses. These are especially
neighboring species in the periodic system of elements, for example,
AsCl3/SeCl3+.
(e) Theoretical calculation of the vibrational modes but this is outside the scope
of this book.
When the experimentally observed data cannot be reasonably assigned to the
possible fundamental modes a new structure must be postulated and the procedure
described must be repeated.
Challenge 2.4
Let us turn again to a molecule of type AB3. The infrared absorption bands
and Raman lines listed in Table 2.3 are gathered from the respective spectra
for the ions SO32� and NO3
�.
What structure do both ions possess?
Solution to Challenge 2.4
The ion of type AB3 can possess a triangular planar structure D3h or a
pyramidal structure C3v. The symmetry of the fundamental modes and the
results of the selection rules are already listed in Table 2.2. As Table 2.2
shows the D3h and C3v structures can unambiguously be distinguished just on
grounds of the number of the experimentally observed infrared absorption
bands and Raman lines. All modes must be infrared and Raman active for the
C3v structure. This is realized for SO32�. The ion NO3
� belongs to the D3h
structure because only three infrared absorption bands and three Raman lines
can be observed in the respective spectra.
But the justification of the proposed structures requires the assignment of
all experimentally observed data. The two highest wave numbers must be
assigned to the two stretching vibrations in the ions according to the rules
given above. The polarized Raman line belongs to the symmetrical stretching
vibration, thus the other one has to be assigned to the asymmetrical one. In the
same manner, the symmetrical bending modes can be assigned by means of
(continued)
Table 2.3 Experimental IR-absorption bands and Raman lines of the ions SO32� and
NO3� given in cm�1 (p polarized Raman line)
SO32� IR 966 935 622 470
Raman 967 (p) 933 620 (p) 469
NO3� IR 1,370 828 695
Raman 1,390 1,049 (p) 716
2.2 Interpretation of Vibrational Spectra of Small Molecules and Ions 79
the polarization of the Raman lines. The reasonable assignment of all wave
numbers summarized in Table 2.4 justifies the assumed structures for both
ions.
Vibrational spectra of materials in the solid state are determined by the symme-
try of the crystal. When the symmetry of the crystal is lower than that of the free ion,
then the following alterations are observed:
1. Splitting up of degenerate vibrations
2. Occurrence of modes that are symmetry-forbidden in the free ion.
Furthermore, interionical and intermolecular interactions in the crystal lattice
cause shifts in comparison with free molecules and ions, respectively and additional
translational and rotational modes of the lattice components appear in the lower
frequency region (n < 300 cm�1) and overtones as well as combination vibrations
are observed in the general middle region. Thus, the spectra of solids are mostly
characterized by more infrared absorption bands than observed for the free mole-
cule and ion. The selection rules for vibrational spectra of solids are determined by
the site-symmetry (see specialized literature). Band-rich infrared absorption spectraare provided by many inorganic salts with highly symmetrical ions, such as sulfates
or carbonates.
Challenge 2.5
Note that the solutions can be found in extras.springer.com.The sign “p” means that the respective Raman line is polarized and “Ra” is
the abbreviation for Raman.
(continued)
Table 2.4 Selection rules and assignment of the infrared absorption bands and Raman
lines of the experimentally observed wave numbers (cm�1) to the theoretically possible
vibrations of the ions SO32� and NO3
�
SO32� NO3
�
C3v IR Raman Assignment
(IR/Raman)
D3h IR Raman Assignment
(IR/Raman)
Stretching vibrations Stretching vibrations
ns(A1) + + 966/967 ns(A10 ) � + –/1,049
nas(E) + + 935/933 nas(E0) + + 1,390/1,390
Bending vibrations Bending vibrations
ds(A1) + + 622/620 g(A2
00) + – 828/–
das(E) + + 470/469 d(E0) + + 695/716
1. The structure of the molecules and ions is to be determined with the
presented data (given in cm�1) gathered from the infrared and Raman
spectra following the steps explained above.
80 2 Vibrational Spectroscopy
(continued)
(a) BF3 Ra 1,506 888 p 482
IR 1,505 719 480
(b) NO2� Ra 1,323 p 1,296 827
IR 1,323 1,296 825
(c) NO2+ Ra 1,296 p
IR 2,360 570
(d) VOCl3 Ra 1,035 p 504 409 p 249 164 p 131
IR All vibrations are infrared active
(e) AlCl4� Ra 498 348 p 182 119
IR 496 FIR: 180
(f) SO3(g) Ra 1,390 1,065 p 530
IR 1,391 529 495
(g) AsF5 Ra 809 733 p 642 p 388 366 123
IR 810 784 400 365 FIR: 122
(h) S2O32� Ra 1,123 995 p 669 p 541 446 p
IR 1,122 995 667 543 445
(i) HSO3� Ra 2,588 p 1,200 1,123 1,038 p 629 p 509
IR 2,585 1,201 1,122 1,038 628 510
2. The stretching modes of [PtCl4]2� are (given in cm�1):
Ra 330 p 312
IR 313
Explain why this ion cannot have a tetrahedral structure? What struc-
ture can be proposed from these data?
3. Determine the structure of the ion [AsF4]� and assign the infrared
absorption bands and Raman lines (given in cm�1):
Ra 829 745 p 272 213
IR 830 270
4. Deduce the structure of PCl5 in the gaseous and solid state and assign all
infrared absorption bands and Raman lines (given in cm�1).Solid state
Ra 662 458 p 360 p 283 255 238 178
IR 661 444 285 254
Gaseous state and in a nonaqueous solution
Ra 580 392 p 281 p 272 261 102
IR 583 443 300 272 101
2.2 Interpretation of Vibrational Spectra of Small Molecules and Ions 81
(continued)
5. Derive the structure of PF5 from the data obtained by the infrared and
Raman spectra (given in cm�1). Assign all infrared absorption bands andRaman lines.
Ra 1,026 817 p 640 p 542 514 300
IR 1,025 944 575 533 302
6. Develop a proposal for the structure of the ion O2PH22� based on the data
(given in cm�1) gathered from the Raman spectrum. Assign the Raman
lines to the theoretical modes if this is unambiguously possible. Note that
the vibration that corresponds to the Raman line at 930 cm�1 is infraredinactive.
Ra 2,365 p 2,308 1,180 1,160 p 1,093 1,046 p 930 820 470 p
7. What structure can be derived from the data of the Raman spectrum
(given in cm�1) for NSF3? Note that all vibrations are also infrared
active. Assign the Raman lines to the corresponding modes of the
molecule.
Ra (liquid) 1,523 p 815 773 p 525 p 432 346
8. The reaction of PCl5 with AlCl3 gives rise to a product for which the
atomic ratio Al : P : Cl ¼ 1 : 1 : 8 is analytically obtained. The structure
is to be derived on the basis of the data gathered from the infrared and
Raman spectra (given in cm�1). Assign all bands and lines.
Ra 662 498 458 p 348 p 255 182 178 120
IR 665 497 258 FIR: 181
9. Derive the structure of an arsenic compound with the sum formula
AsCl2F3 and assign the infrared absorption bands and Raman lines
(given in cm�1).
Ra 689 p 573 500 422 p 375 187 157
IR 700 500 385 FIR: 186
10. The addition of a solution of SeCl4 in dioxane to a solution of AlCl3 in
dioxane gives rise to a solid product for which the atomic ratio Al : Se :
Cl ¼ 1 : 1 : 7 is analytically obtained. Derive the structure of this reaction
product from the data of the infrared and Raman spectra (given in cm�1).Assign the infrared bands and Raman lines. Note that KAlCl4 shows lines
at 182 and 119 cm�1 in the region n < 200 cm�1 in the Raman spectrum.
Ra 498 416 p 395 348 p 294 p 186 182 119
IR 497 415 396 296 FIR: Not detected
82 2 Vibrational Spectroscopy
(continued)
11. A platinum compound has the analytical composition PtCl4•2H2O•2HCl.
Develop a proposal for the structure of this solid reaction product on the
basis of the infrared absorption bands and the Raman lines (given in cm�1).Assign the corresponding bands to the theoretical modes.
Ra 3,226 p 2,825 1,695 1,070 p 348 p 318 171
IR 3,225 2,825 1,694 1,068 342 183
12. Compose a table with the results of the selection rules for the infrared
absorption bands and Raman lines of the fundamental modes of the struc-
tural moiety X–CH3. Assign the experimentally observed data for X ¼ Cl
(given in cm�1).
Ra 3,041 2,966 p 1,456 1,355 p 1,017 734 p
IR 3,042 2,967 1,455 1,354 1,015 732
Which modes will provide relatively constant values of the wave num-
bers at variation of X and which modes will show a great range for the
observed wave numbers? Explain your findings.
13. Derive the structure of the two compounds OXCl2 with X ¼ C and X ¼ S,
based on the data of the vibrational spectra. Assign the infrared absorption
bands and Raman lines (given in cm�1) to the respective modes of each
molecule.
(a) X ¼ C
Ra 1,827 p 849 580 569 p 440 285 p
IR 1,826 849 568 441 284
(b) X ¼ S
Ra 1,251 p 492 p 455 344 p 284 194 p
IR 1,250 490 454 342 284 FIR: Not detected
14. The totally symmetrical vibration n1 in the fluoric complexes of the third
main group of the type [MeF6]3�with Me ¼ Al, Ga, In, Tl is observed in the
Raman spectrum at 478, 498, 535, and 541 cm�1, respectively ordered
according to increasing wave numbers. Assign n1 to the respective Me–X
mode. Why is the totally symmetrical vibration at best suitable in order to
evaluate the bond strength via the force constant?
15. The totally symmetric vibration n1 in the fluoric complexes of vanadium of the
type [VF6]n� with n ¼ 1, 2, and 3 is observed at 533, 584, and 676 cm�1,
respectively. Assign n1 to the respective complex ion and justify your decision.
16. Perovskites of the type A2BMO6 can be divided according to their [MO6]-
structure moiety into three groups:
(a) Perovskites with an undistorted [MO6]-octahedron
(b) Perovskites with two different undistorted [MO6]-octahedrons
(c) Perovskites with a distorted [MO6]-octahedron.
2.2 Interpretation of Vibrational Spectra of Small Molecules and Ions 83
(continued)
800600400200
p
a
p
b
p
c
200 400 600 800
200 400 600 800
n in cm-1
n in cm-1
n in cm-1
p
Fig. 2.7 Line diagram of the
Raman spectrum of three
perovskites. “p” means that
the respective Raman line is
polarized
Irel
I||
314
400300200100
I⊥ 368
8
n in cm-1
Fig. 2.8 Raman spectra of
the [FeS4] structural moiety in
a peptide obtained by
polarized light
Assign the line diagrams of the Raman lines for three perovskites shown in
Fig. 2.7 to the three types of perovskites and justify your decision.
17. The [FeS4] structure moiety of a ferric-containing peptide can be planar or
tetrahedral. Explain why Raman spectroscopy must be applied for solving this
problem. The Raman spectra obtained by polarized light are presented in
Fig. 2.8. What structure has the [FeS4] moiety? Assign the stretching
84 2 Vibrational Spectroscopy
2.2.2 Coordination of Bifunctional Ligands at the CentralAtom in Complex Compounds
The thiocyanate ion SCN� is a bifunctional ligand because it can be coordinated viathe S and the N atom to a central atom in a complex compound. Vibrational
spectroscopy can help to distinguish between both structures and this is explained
in the following.
A linear three-atomic ion provides 3 � 4–5 ¼ 4 fundamental modes, which are
divided into two stretching and two bending vibrations. But the infrared spectrum
of the free SCN� ion only has three absorption bands because the two bending
modes at 470 cm�1 are degenerated. The infrared absorption bands at 2,053 cm�1
and that at 748 cm�1 can be roughly assigned to n(C�N) and n(C–S), respectively.But the “real” n(C�N) obtained for nitriles lies in the range from 2,260 to
2,240 cm�1 and the “real” C–S stretching vibration is observed at approximately
680 cm�1, see Sect. 2.3. This means, the C�N bond is weaker than a threefold bond
and the C–S bond is stronger than a single bond; thus, the chemical structure must
be described by two resonance structures I and II, respectively (see Structure 2.1):
S C N(I) S C2.1
N(II)
vibrations. Note the empirical fact that ns > nas is valid for the D4h
structure, for the tetrahedral structure the reverse relation is observed.
18. The two degenerated bending vibrations of the free SCN ion are observed
at 470 cm�1 in solutions as an asymmetrical absorption band. The same
mode is split into two absorption bands at 486 and 471 cm�1 in the
infrared spectrum of solid KSCN. What is the reason behind this?
19. There are two structures for CaCO3: Calcite and aragonite. The data
obtained from the vibrational spectra in the region above 400 cm�1 arelisted in Table 2.5.
Assign the infrared absorption bands and Raman lines. Compare qualita-
tively both polymorphic forms of CaCO3.
Table 2.5 Infrared absorption bands and Raman lines of the polymorphic forms of CaCO3:
Calcite and Aragonite
Calcite
IR �1,460 (broad) 879 706
Ra 1,432 1,087 714
Aragonite
IR 1,504 1,492 1,080 866 711 706
Ra The lines correspond to the IR absorption bands.
2.2 Interpretation of Vibrational Spectra of Small Molecules and Ions 85
Coordination to the S atom favors the resonance structure II in Structure 2.1:
Me S�C � N$ Me S ¼ C ¼ N:
The wave numbers of n(C–S) should be smaller and that of n(C�N) should be
higher in a complex of the type Me S–C�N in comparison to the free ligand.
However, coordination to the N atom favors resonance structure I in Structure 2.1:
Me N ¼ C ¼ S$ Me N � C�S:
The wave number of n(C–S) should be increased and the wave number of
n(C�N) should be decreased in a complex of the type Me N¼C¼S in comparison
to the free ligand.
Deviations from these rules can be caused by coupling of vibrations.
Challenge 2.6
The stretching modes of the SCN� ligand in two thiocyanate complexes are:
HgðSCNÞ4� �2�
: 2120 710 cm�1 FeðSCNÞ6� �3�
: 2055 828 cm�1
Judge whether the ligand coordinates via S or N to the central atom.
Solution to Challenge 2.6
HgðSCNÞ4� �2�
Because the wave number of n(C�N) for the complex is higher than that
for the SCN� ion and the wave number of n(C-S) for the complex is smaller
than that for the SCN� ion the coordination occurs via S: Hg SCN.
FeðSCNÞ6� �3�
Because the following facts are valid: n(C�N) (complex)<n(C�N)(SCN�) and n(C-S) (complex)>n(C-S) (SCN�) the coordination occurs via
N: Fe SCN.
86 2 Vibrational Spectroscopy
2.2.3 Investigations of the Strength of Chemical Bonds
Vibrational spectroscopy offers the possibility to investigate chemical bond
strengths because the stretching modes are determined by the force constant
according to Eq. 2.1 which is proportional to the bond order. This fact will be
explained employing the example of nitrate complexes [Me(NO3)n]m�.
The free ligand NO3� shows two stretching vibrations, the symmetrical ns(N-O)
which is infrared inactive and the degenerated asymmetrical nas(N-O) which is
observed at 1,390 cm�1 in the infrared spectrum. Coordination of the ligand to the
metal ion via O diminishes the D3h symmetry of the free NO3� ion with the result
that the degenerated stretching mode is split. Furthermore, the symmetrical
stretching mode is no longer forbidden. Thus, the following stretching modes are
infrared active for a metal complex with a structure moiety shown in Structure 2.2
nðN� OÞ ¼ n1; ns(NO2Þ ¼ n2; and nas(NO2Þ ¼ n3:
Me
ON
O
O2.2
The equality of the three N�O bonds in the free NO3� ion is no longer present
because of the metal-oxygen bond Me O–NO2. Thus, the chemical bond in the
ligand can be described by a (mainly) single N–O bond and two equal N¼O bonds
with conjugated double bonds which provide strongly coupled symmetrical,
ns(NO2) and asymmetrical, nas(NO2) modes, respectively. Because the delocaliza-
tion of the electrons is diminished to only two bonds the strength of the chemical
N¼O bond and hence, the wave numbers of the stretching modes of the NO2 group
increase more, the stronger the bond of the ligand to the metal ion is. In general, the
mean value of the frequencies of the symmetrical and asymmetrical NO2 stretching
modes is used to qualitatively evaluate the strength of the chemical bond of the
ligand to the metal ion. Furthermore, the splitting up of the degenerate N¼Ostretching mode will be all the greater the stronger the complex bond is. The
maximum value is achieved at a “real” covalent bond, i.e., if the metal ion is
substituted by H or an alkyl group; see, for example, the vibrational frequencies of
HNO3 in Table 2.6.
The comparison of stretching modes of the free ligand with those of a compound
possessing a covalent bond of the ligand and those of the complex compound
enables the evaluation of the strength of the chemical bond of the ligand to the
central atom.
2.2 Interpretation of Vibrational Spectra of Small Molecules and Ions 87
Challenge 2.7
Table 2.6 presents the stretching modes of two nitrate complexes and HNO3.
Decide whether NO3� coordinates to each of the metal ions. If this is
correct, evaluate qualitatively the strength of the Me O– NO2 bond.
Solution to Challenge 2.7
The degeneration of the asymmetrical stretching vibration of the free NO3� ion
is no longer valid. Instead it is split into the symmetrical and asymmetrical NO2
stretching vibrations. Furthermore, IR-forbidden symmetrical stretching is
allowed in both complexes. Thus, the D3h symmetry is no longer given because
of coordination of the NO3� ligand to the metal ions.
To qualitatively evaluate the coordinative bond strength the mean values
calculated by the wave number of the symmetrical and asymmetrical N�Ostretching modes are compared with that of HNO3 in which the H�O bond is
a “true” covalent one. As the wave numbers in the last column of Table 2.6
show the coordinative bond is stronger in the Zr complex than in the Cu one
because the mean value of the NO2 stretching modes of the Zr complex is
higher than that of the Cu complex and is closer to the value for a strong
covalent bond.
Note that the procedure for the qualitative evaluation of the strength of a
coordinative chemical bond described above can only be applied with true
characteristic vibrations whose vibrational frequencies are influenced solely
by electronic effects and not by couplings of vibrations.
Challenge 2.8
For solutions to Challenge 2.8, see extras.springer.com.
1. K3[Co(CN)5(NCS)] can appear in two isomeric forms which give rise to
the following infrared absorption bands:
Isomer I, n in cm�1 2,065 810 Isomer II, n in cm�1 2,110 718
Draw up and justify a proposal of the structure for the two isomeric forms
of the Co(III) complex.
(continued)
Table 2.6 Stretching modes of two nitrate complexes and HNO3
n1 n2 ¼ ns(NO2) n3 ¼ nas(NO2) Dn ¼ n3�n2 �n(n2, n3)
[Cu(NO3)4]2� 1,013 1,290 1,465 175 1,377.5
[Zr(NO3)6]2� 1,016 1,294 1,672 378 1,428.5
HONO2 920 1,294 1,672 378 1,483
88 2 Vibrational Spectroscopy
2. The following infrared absorption bands are observed for the complex Pd
(bipyr)(NSC)2 in the region of the NCS stretching modes:
2,117 2,095 842 700 cm�1
Draw up and justify a proposal of the coordinative chemical bond in this
complex compound.
3. Figures 2.9 and 2.10 show sections of the infrared spectra of the selenium
compounds (C6H5)2Se(NO3)2 and (C6H5)3SeNO3, respectively. Investi-
gate whether nitrate is preferably covalently bonded to the selenium atom
or whether an ionic structure with an NO3� ion is valid.
4. The mean values of the symmetrical and asymmetrical stretching
vibrations of NO2, NO2+, and NO2
� ordered according to increasing
wave numbers are: 1,307, 1,468, and 1,878 cm�1. Assign and justify the
wave numbers to the corresponding nitrogen compounds.
5. Evaluate the strength of the coordinative chemical bond of the oxalate
ligand to the metal ions Pt and Cu on the basis of the C¼O stretching
vibrations. The observed infrared absorption bands of these complex
compounds as well as those of the noncoordinate, free oxalate ion and
covalently bonded oxalic acid dimethyl ester are summarized in Table 2.7.
6. The stretching frequencies of manganese compounds of the type KnMnO4
with n ¼ 1, 2, and 3 are arbitrarily listed in Table 2.8. Assign the infrared
absorption bands and Raman lines to the respective manganese
compounds. Explain why one can recognize a relation between the wave
(continued)
A in %
1600140012001000800
100
0
n in cm-1
Fig. 2.9 Infrared spectrum in
the range of the NO stretching
modes of (C6H5)2Se(NO3)2;
sampling technique: Nujol
mull
2.2 Interpretation of Vibrational Spectra of Small Molecules and Ions 89
number of the symmetrical stretching modes and the effective charge of
the Mn atom?
7. The CO stretching vibration of C�O can be found at 2,155 cm�1 but it liesin the range from 2,100 to 1,800 cm�1 in metal carbonyl compounds. The
chemical bond in metal carbonyl compounds is a superposition of
a s-Me-C bond and a Me-(CO)-p back donation. What information is
obtained by the mean values of the C¼O stretching modes according to the
part of both contributions to the chemical bond in the following charged
(continued)
HCBD
A in %
100 in Nujol in HCBD
1200 1400 160010008000
n in cm-1
Fig. 2.10 Infrared spectrum in the range of the NO stretching modes of (C6H5)3Se(NO3);
sampling techniques used: Nujol mull (left) and hexachlor-butadiene (HCBD) (right)
Table 2.7 C¼O stretching frequencies of the oxalate ion, oxalic acid dimethyl ester, and
some oxalate complexes (given in cm�1)C2O4
2� 1,640 1,650 C2O4(CH3)2 1,770 1,796
[Pt(C2O4)4]2� 1,674 1,709 [Cu(C2O4)4]
2� 1,645 1,672
Table 2.8 Arbitrarily ordered infrared absorption bands and Raman lines of manganese
compounds of the type KnMnO4
Compound I: Raman: 820 812 332 325 cm�1
IR: 821 333 cm�1
Compound II: Raman: 902 834 386 346 cm�1
IR: 901 384 cm�1
Compound III: Raman: 789 778 332 308 cm�1
IR: 779 331 cm�1
90 2 Vibrational Spectroscopy
metal carbonyl ions? The wave number of Ni(CO)4 may be used as the
reference compound. All values are given in cm�1.
Ni(COÞ4ð2094Þ; Co(COÞ4� ��ð1946Þ; Fe(COÞ4
� �2�ð1788Þ; Pt(COÞ42þð2258Þ:
2.3 Interpretation of Vibrational Spectra of Organic Molecules
2.3.1 Objectives and Special Features
The interpretation of vibrational spectra mainly includes the following objectives
• Determination of structural moieties (alkyl, alkenyl-, alkin-, and aryl groups)
• Determination of functional groups• Determination of isomers• Investigation of conformers• Investigation of hydrogen bridges (which is only possible by means of infrared
spectroscopy)
In contrast to the interpretation of vibrational spectra of inorganic species
symmetry properties of modes are, generally, not of interest.
The occurrence of overtones and combination vibrations increases whereas ran-
domly degenerated vibrations decrease the number of the 3 � N–6 possible modes.
Unlike the degeneration of modes on the basis of symmetry, randomly degeneratedvibrations are caused if two or more atomic groups accidently possess the same
vibrational frequency. This is the case, for example, for acetone, CH3–C¼O–CH3.
The inner vibrations of the two CH3 groups accidently possess the same vibrational
frequencies resulting in one symmetrical ns(CH3), (degenerated) asymmetrical
nas(CH3) stretching modes, symmetrical ds(CH3), and (degenerated) asymmetrical
das(CH3) bending modes, respectively for both CH3 groups. Whereas almost allinfrared absorption bands and Raman lines can be assigned to the respective
vibrations for small inorganic species the assignment is restricted only to a few for
organic molecules. The extensive coupling of vibrations of the atoms of similar
masses (C, O, N) linked with similar bond strength (C–C, C–O, C–N) give rise to
many absorption bands whose source is mostly unknown. Thus, such absorption
bands cannot be assigned to the vibration of a certain atomic group.
The following vibrations are meaningful for the interpretation of the vibrational
spectra of organic molecules:
• Characteristic vibrations
• Structure-specific coupled vibrations
• The fingerprint region
2.3 Interpretation of Vibrational Spectra of Organic Molecules 91
2.3.2 Characteristic Vibrations
2.3.2.1 Definition
Characteristic vibrations of a multiatomic molecule are present if nearly the wholepotential energy is localized in the movement of a certain atomic group of themolecule with the consequence that coupling of vibrations is negligible. Thus,characteristic vibrations give rise to infrared absorption bands at about the samewave numbers, regardless of the structure of the rest of the molecule.
These site-stable infrared absorption bands are called group wave numbers (alsocalled group frequencies) because the absorption band can be unambiguously
assigned to the vibration of certain atomic groups; therefore, they provide informa-
tion about the absence or presence of specific atomic groups. Thus, they are
valuable diagnostic markers for the recognition of functional groups. Group wave
numbers are listed in tables and they are the basis for vibrational spectra
interpretation.
2.3.2.2 Criteria for Characteristic Vibrations
Characteristic vibrations can occur if
1. The force constants of neighbored bonds are different (greater 25%).
2. The mass of neighbored atoms are strongly different (greater 100%).
3. The vibration is not accidently in the same region as another vibration.
The first and second criteria are very well fulfilled by, for example, H–C�N.Infrared radiation at 3,311 cm�1 excites the vibration of only the H–C group
because, according to calculations, 95% of the absorbed energy is utilized for the
movement of this atomic group. However, 95% of the absorbed energy at
2,097 cm�1 is used for the vibration of the C�N group. Because both stretching
vibrations fulfill the criteria for characteristic vibrations the region between 3,300
and 3,320 cm�1 and at �2,100 cm�1 are the group wave numbers for the C–H
stretching vibration with a sp-hybridized C atom and the C�N stretching one,
respectively. Note that the group wave numbers of the C�N and C�C groups are
nearly the same because of the similar bond order and the approximately similar
reduced masses.
According to the first criterion the stretching vibrations of nonconjugated multi-
ple bonds provide characteristic vibrations whose group wave numbers are diag-
nostically valuable for the unambiguous recognition of the structural groups C¼C,C�C, C¼O, C¼N, or C�N.
The first criterion is also fulfilled for the C¼S group. Therefore, the C¼S group
should provide a characteristic vibration and the wave number range of this group
should be consistent in analogy to the C¼O group. Unfortunately this is not the
case. Because of the greater mass of S the stretching frequency is diminished and
92 2 Vibrational Spectroscopy
lies in the range of the bending vibrations of the CH skeleton. Thus, criterion 3 is
not fulfilled and the C¼S stretching vibrations cannot be assumed to be a charac-
teristic vibration with a small range of group wave numbers. According to
calculations a large amount of the absorbed energy of the expected range of the
C¼S stretching vibration is distributed to bending vibrations of the CH skeleton, in
other words, the C¼S stretching vibration is a strongly coupled one and the
denotation n(C¼S) for an intense absorption band in the relatively large range of
the C¼S stretching vibrations is only approximately valid.
Because, in general, low-energy bending vibrations are strongly coupled the
presence of characteristic vibrations is limited to stretching vibrations except for the
inner bending vibrations of the CH3 and CH2 group.
The respective inner vibrations of the CH3 and CH2 groups are characterized by
the fact that the vibration is mainly limited to the movement of CH atoms whereas
outer vibrations also include movements of the neighbored atoms which are named
rocking (r), wagging (o), and twisting (t). It is clear that these vibrations are not
characteristic and do not provide group frequencies. Therefore, they are not of
interest for structural analysis.
2.3.2.3 Examples of Characteristic Vibrations (Group Wave Numbers)
According to Criterion 1 (f1 6¼ f2)
Examples are listed in Table 2.9.
2.3.2.4 Examples of Characteristic Vibrations (Group Wave Numbers)
According to Criterion 2 (m1 6¼ m2)
Examples are listed in Table 2.10.
2.3.2.5 Internal and External Influences on Characteristic Vibrations
Because of negligible coupling of characteristic vibrations their group wave numbers
should be observed in a narrow range. But this is not the case as is shown, for
Table 2.9 Examples of group wave numbers (in cm�1) of compounds with multiple bonds
C C X
f1 f
2X ¼ O Carbonyls n(C¼O) � 1,850–1,650
X ¼ N Azomethine n(C¼N) � 1,650
X ¼ C Alkene n(C¼C) � 1,640
C C
f1 f2X
X ¼ N Nitrile n(C�N) � 2,260–2,200
X ¼ C Alkyne n(C�C) � 2,260–2,100
2.3 Interpretation of Vibrational Spectra of Organic Molecules 93
example, for the group wave numbers of the C–H stretching vibrations (see
Tables 2.10). The reason for this fact is that characteristic vibrations are influenced
by internal and external effects. Knowledge of the correlation between peak position
and these effects is valuable to recognize functional groups and structural moieties.
2.3.2.6 Inductive Effects
Increasing of the –I effect increases the bond order and, hence, the force constant in
carbonyl compounds. Therefore, the wave number of the C¼O stretching vibrations
increases with the strength of the –I effect of X which is shown by the following
examples:
CH3(C¼O)–X X ¼ H n(C¼O): 1,740 cm�1 X ¼ Cl n(C¼O): 1,800 cm�1
2.3.2.7 Mesomeric Effects
Mesomerism diminishes the bond order and, hence, the force constant resulting in
lower wave numbers of the stretching vibration. Examples are:
CH3(C¼O)–X X ¼ H n(C¼O): 1,740 cm�1
X ¼ Aryl n(C¼O): 1,685 cm�1
–C¼C–X X ¼ H n(C¼C): � 1,645–1,640 cm�1
X ¼ Aryl n(C¼C): � 1,600–1,450 cm�1
2.3.2.8 Intermolecular Effects
Autoassociation, hydrogen bridges, interactions with the solvent and further effects
diminish the strength of the chemical bond and, hence, the respective stretching
vibrations are observed at lower wave numbers as is demonstrated by the
Table 2.10 Examples of group wave numbers (in cm�1) of compounds with the –C–X–H
structural moiety
C X H
m1 m
2X ¼ O Alcohol, phenol n(OH)free � 3,650
Not associated n(OH)ass � 3,200
Associated n(NH) � 3,400
X ¼ N Amine, amide 2 IR-absorption bands
Primary 1 IR-absorption band
Secondary n(CH) � 3,300–2,850
X ¼ C Hydrocarbon
�C(sp)–H n(CH) � 3,300
¼C(sp2)–H n(CH) � 3,100–3,000
–C(sp3)–H n(CH) � 3,000–2,850
94 2 Vibrational Spectroscopy
experimentally observed wave numbers of the C¼O stretching vibration of acetone:
n(C¼O), gaseous: 1,745 cm�1, n(C¼O), in CCl4: 1,720 cm�1. Therefore, informa-
tion on the preparation techniques used for the registration of the vibrational
spectrum of a sample should always be given.
2.3.3 Structure-Specific Coupled Vibrations
In contrast to characteristic vibrations structure-specific coupled vibrations cannot
be assigned to the vibration of certain atomic groups but they appear as a known
absorption band pattern in the infrared spectrum which is specific for a certain
structural moiety.
Structure-specific coupled vibrations are accompanied by the presence of cer-tain structural moieties in the molecule. Therefore, they diagnose a certain struc-tural type rather than a functional group in a molecule.
Examples of structure-specific coupled vibrations
1. Overtones and combination vibrations of the CH out-of-plane bending modes
and ring bending modes of the CC skeleton of aromatic compounds result in
absorption bands in the range from 2,000 to 1,600 cm�1 whose pattern enables
the recognition of the substitution type of the aromatic compound by comparison
of the absorption band pattern with those of “ideal” ones shown in Fig. 6.1. Note
that these absorption bands give rise to weak intensities. Hence, in general, they
can be observed as more intense bands at higher concentrations or layer thick-
ness of the cuvette.
2. The two intense (sometimes only one) infrared absorption bands in the range
from 2,850 to 2,700 cm�1 diagnose the presence of the formyl group CHO in the
molecule. As explained above these absorption bands specific for a formyl group
are caused by Fermi resonance.
3. The weak infrared absorption band in the range from 1,840 to 1,800 cm�1 due toa combination of the two out-of-plane bending vibrations at �990 and �910cm�1 diagnose the vinyl-group, –CH¼CH2.
2.3.4 The Fingerprint Region
The absorption bands in the range <1,500 cm�1 mostly result from strongly
coupled vibrations that cannot be assigned to movements of certain atomic groups.
Therefore, their position in the spectrum is determined by a certain structure of the
molecule and only small structural changes in the molecule will markedly alter the
pattern of these coupled vibrations. Whereas characteristic vibrations provide group
wave numbers for functional groups and structure moieties regardless of the
structure of the rest of the molecule the strongly coupled vibrations with absorption
2.3 Interpretation of Vibrational Spectra of Organic Molecules 95
bands <1,500 cm�1 are specific for a single molecule. Therefore, this region is
named the fingerprint region.Whereas characteristic vibrations diagnose only functional groups and struc-
ture moieties the fingerprint region facilitates the identification of an individualcompound.
The identification of substances is realized by visual or computer-aided compar-
ison of the infrared absorption spectrum of the sample with that of a reference
sample and a reference spectrum from a spectroscopic library, on condition that
both spectra were measured under the same conditions. Especially, the pattern of
the fingerprint region is valuable to recognize an individual compound crystallized
in a special lattice. Bands of residual solvents or the habit of the crystal can give rise
to additional, but weakly intense bands and slightly change the position of bands in
the spectrum, respectively. But the most intense absorption bands may not differ
more than �1 cm�1.Additional infrared absorption bands or Raman lines or changed band positions,
especially observed in the low wave number range, can be caused by different
crystal structures, i.e., polymorphic forms. Most of the organic compounds crystal-
lize in polymorphic forms. Because vibrational spectra can be easily obtained from
solids, vibrational spectroscopy is an excellent and simple method in order to
determine polymorphic structures in routine analysis. Thus, such investigations
are included, for example, in the regulations (Pharmacopeia) for pharmaceutical
analysis. Note that polymorphic forms differ in physical properties such as solubil-
ity and stability. Therefore, the determination of the correct polymorphic form is an
important condition to apply a compound as an active pharmaceutical ingredient.
Besides vibrational spectroscopy, X-ray diffraction, thermal analytical methods,
photoacoustic spectroscopy, and solid-state NMR spectroscopy are further methods
to investigate polymorphism.
Because Raman spectroscopy, in contrast to routine infrared spectroscopy, also
allows recording of the low-frequency range of lattice vibrations which are valuable
for recognition of polymorphic forms, Raman spectroscopy is an excellent method
for the investigation of polymorphism whereas Raman microscopy enables the
investigation of single crystallites. But thermal transformation of polymorphic
forms during the measurement must be considered.
2.3.5 Interpretation of Vibrational Spectra of Hydrocarbons
Vibrational spectra provide information about
1. The structural type of hydrocarbons (alkanes, alkenes, alkynes, aromatic
hydrocarbons),
2. Branching of the carbon-carbon chains,
3. Structures of branch points,
4. Bonding geometry around the C¼C-bond of alkenes,
5. Substitution pattern of aromatic compounds.
96 2 Vibrational Spectroscopy
The most useful diagnostic bands to determine these structural alterations are the
C–H, C¼C, and C�C stretching vibrations, the internal bending vibrations of the
CH3 and CH2 groups, the out-of-plane bending modes of alkenes and aromatic
compounds g(CH) or doop, the out-of-plane bending modes of the CC skeleton of
the aromatics and their over and combination tones which are summarized in
Tables 2.11, 2.12 and 2.13 and in detail in Table 6.15.I. Note that the din-plane ofalkenes and aromatics are meaningless for structural determinations.
Challenge 2.9
Figure 2.11 shows the infrared absorption spectrum of a synthetic product.
Justify that the spectrum can be assigned to para-methyl styrene (Structure 2.3).
(continued)
Table 2.11 Diagnostic information (wave numbers in cm�1) for alkanes (sp3C–H)CH stretching vibrations, nCH < 3,000
CH3 nas(E): � 2,965 ns: 2,875 I(nas) > I(ns)CH2 nas: � 2,925 ns: 2,855 I(nas) > I(ns)CH � 2,900 (weak intensity; mostly overlaid)
Deviating position and intensity at CH3�X and CH2�X with X ¼ O, N, Aryl
Bending vibrations
CH3 das(E) � 1,470 ds � 1,375 Splitting up at branching
CH2 d � 1,455
Table 2.12 Diagnostic information (wave numbers in cm�1) for alkenes and aromatics (sp2C–H)
CH stretching vibration, nCH < 3,000
Alkenes Aromatics
n(C¼C) 1,690–1,635 n(C¼C) 1,625–1,400
>C¼C < types Aryl substitution types
�CH¼CH�(E/Z) g(CH) g(CH) + ring bending at mono
and meta substitution
850–650
�CH¼CH2 1,005–675 Only g(CH) at ortho and parasubstitutions
2,000–1,000 Band pattern,
see Fig. 6.1
>C¼CH2 Overtones and combination
vibrations
Table 2.13 Diagnostic information (wave numbers in cm�1) for alkynes (spC–H)CH stretching vibration, nCH 3,340–3,250
Sharper than n(OH) and n(NH)C�C stretching vibration, nC�C 2,260–2,100
Intensity is small to missing
Bending vibration, dC�CH n � 650 with an overtone at n � 1,250
2.3 Interpretation of Vibrational Spectra of Organic Molecules 97
CH=CH2
CH3 2.3
Solution to Challenge 2.9
The structural moieties that can be spectroscopically justified by infrared
spectroscopy are summarized in Table 2.14.
1886 1805
1609(sh)
1486
2864
T in %
0
50
30683048 825
1628
1513
1450
2922 + 2925 (sh)
990
905
1000 50015003000 200025003500
n in cm-1
1570
Fig. 2.11 Infrared spectrum of the synthetic product para-methyl styrene (Structure 2.3);
sampling technique: capillary thin film
Table 2.14 Assignment of the IR-absorption bands to para-methyl styrene (Structure 2.3)
Structural moiety Wave numbers in cm�1 Assignment of the infrared absorption bands
Aromatic
compounds
3,070–3,010 n(sp2C–H)1,570 w, 1,513 s,
1,486 w
n(sp2C¼C), Aryl
–CH¼CH2 3,070–3,010 n(sp2C–H)1,628 m n(C¼C), in conjugation
990 s + 905 s g(CH) for –CH¼CH2
CH3–aryl 2,922 m + 2,864 m ns(CH3) + Fermi resonance, Aryl–CH3
2,925 (sh), w nas(CH3) for aryl–CH3
para-substitution 825 s g(CH), Aryl(No ring bending!)
2,000–1,650 Pattern of overtones and combination
vibrations
98 2 Vibrational Spectroscopy
Challenge 2.10
For solutions to Challenge 2.10, see extras.springer.com.
1. Figures 2.12, 2.13, 2.14 and 2.15 present infrared spectra of hydrocarbons.
Assign the infrared absorption bands to the vibration of alkanes, alkenes,
alkynes, and aromatic compounds. What structural moieties can and
cannot be reliably recognized? The relevant wave numbers for the struc-
ture determination of the respective compounds are marked in bold.
a
C C
CH3 CH3
CH3
CH3
CH3
CH3 2.4
Assign the following infrared bands:
2,971; 2,875; 1,480; 1,448; 1,380/1,365; 1,180 cm�1
What spectroscopic information gives rise to the evidence for
(1) The absence of a CH2 group,
(2) The presence of the tertiary butyl group?
(continued)
100
29712875
14801180
50
T in %
3000 1500 10002900
14481380 (42 %)1365 (22 %)
n in cm-1
Fig. 2.12 Infrared spectrum of 2,2,3,3-tetra-methyl butane (Structure 2.4); sampling
technique: capillary thin film
2.3 Interpretation of Vibrational Spectra of Organic Molecules 99
bH2C C
CH3
CH2 CH3
2.5
(continued)
100
1780T in %
50
30782970291828662855
16501468
1377
888
1237 1074
2900 1500 10003000
n in cm-1
Fig. 2.13 Infrared spectrum of 2-methyl buten-1 (Structure 2.5); sampling technique:
capillary thin film
100
50
2962
2937
2120
14601432
1380
630
1255
T in %
30003500 2000 10001500
3311 2877
n in cm-1
Fig. 2.14 Infrared spectrum of hexin-1 (Structure 2.6); sampling technique: capillary thin
film
100 2 Vibrational Spectroscopy
Assign the following infrared bands:
3,078; 2,970; 2,918; 2,866; 2,855; 1,780; 1,650; 1,468; 1,377; 888 cm�1.The bold wave numbers belong to the alkene.
What spectroscopic information provides evidence for
(a) The presence of a CH2¼CR1R2 moiety,
(b) An unbranched C–C chain?
cHC C (CH2)3 CH3
2.6
Assign the following infrared bands:
3,311; 2,962; 2,937; 2,877; 2,120; 1,460; 1,432; 1,380; 1,255; 630 cm�1
The bold wave numbers belong to the alkyne.
What spectroscopic information provides evidence for an unbranched C–C
chain?
dCH3
2.7
Assign the following infrared absorption bands:
3,087; 3,062; 3,038; 2,948; 2,920; 2,870; 2,000–1,650; 1,605–1,495;
1,460; 1,380; 728; 694 cm�1
(continued)
100
50 29481605 1460
2870 1380T in %
10003000 15002000
308730623038
29201495
728 694
n in cm-1
Fig. 2.15 Infrared spectrum of toluene (Structure 2.7); sampling technique: capillary thin
film
2.3 Interpretation of Vibrational Spectra of Organic Molecules 101
The bold wave numbers belong to the aromatic part of the molecule.
What spectroscopic information gives rise to the evidence for
(a) The presence of an aromatic and the absence of a C¼C group,
(b) The mono-substitution of the aromatic compound.
(c) The presence of the CH3–aryl structural moiety?
(continued)
a b
50
100
T in %
29552925
2868
2958
2870
2928
30003000
n in cm-1
Fig. 2.16 (a, b) Infrared spectra of hydrocarbons extracted from seepage water; sampling
technique: capillary thin film
50
3000 2500 2000 1500 1000
100
306630503018
2940292128782860
1604
1495
1460
13781080
1030
742
T in %
n in cm-1
Fig. 2.17 Infrared spectrum of a liquid hydrocarbon compound; sampling technique:
capillary thin film. MS : m=zðIrel in%Þ : 106ð68ÞMþ; 107ð6:0Þ
102 2 Vibrational Spectroscopy
2. Seepage water of two retention basins at the interstate was extracted by
C2Cl3F3. The infrared spectra measured in the range of the C–H stretching
vibrations after removal of the extraction solvent are shown in Fig. 2.16a, b.
(continued)
T in %
100a b
3000
50
2500 1500
29602927
28742861
1465
1379
100
3000 2500 1500 1000
50
2928
2853
1452
1257
904852
T in %
n in cm-1n in cm-1
Fig. 2.18 (a, b) Infrared spectra of two C6-hydrocarbon compounds; sampling technique:
(a) in CCl4; (b) capillary thin film
100
3000
50
2500 1500 1000
1658
14671457
13781370
698
2957293028752860
3016
T in %
n in cm-1
Fig. 2.19 Infrared spectrum of a liquid hydrocarbon compound; sampling technique:
capillary thin film. h�MS : m=z ¼ 98:1097 amu Mþ
2.3 Interpretation of Vibrational Spectra of Organic Molecules 103
Determine and justify whether the infrared spectra belong to gasoline or
diesel oil.
3. Figures 2.17, 2.18, 2.19, and 2.20 present the infrared spectra of
hydrocarbons. Determine structural moieties of these compounds and
propose a structure for these with the aid of the additional information
given.
(continued)
100
50
1000150020003000
3312
30392922
28612106
15981581
14841453
1379
785
691
653
T in %
n in cm-1
a
100
50
3000 2500 2000 1500 1000
310030783036 2958
2917
28622250
2217
15991571
14941464 756 692
10721032
996
T in %
b
n in cm-1
Fig. 2.20 (a, b) Infrared spectra of two structurally isomeric hydrocarbon compounds;
sampling technique: capillary thin film. h�MS : m=z ¼ 116:0627 amu Mþ
104 2 Vibrational Spectroscopy
100
50
250030003500
an (OH, non associated)
T in %
n in cm-1
100
50
250030003500
b
n (OH, polymer)
n (OH, non associated)n (OH, dimer)
T in %
n in cm-1
100
50
3500 3000 2500
c
n (OH, polymer)
T in %
n in cm-1
Fig. 2.21 (a–c) Infrared spectra of the O–H stretching vibrations of solutions of n-butanolin CCl4 with various concentrations (in mol L�1) and thicknesses of the cuvettes, l (in mm):
(a) c ¼ 0.005, l ¼ 50; (b) c ¼ 0.125, l ¼ 2 mm; (c) c ¼ 2, l ¼ 0.05 mm
2.3 Interpretation of Vibrational Spectra of Organic Molecules 105
2.3.6 Interpretation of Vibrational Spectra of CompoundsContaining the Functional Group C–X–R withX ¼ O, N, and S
2.3.6.1 Alcohols and Phenols, C–O–H
The diagnostically valuable vibrations are:
H–O stretching vibrationsWave number range:
nðOHÞfree 3650� 3585 cm�1 ðsharp and intenseÞ
nðOHÞassociated 3550� 3200 cm�1 ðvery broad and very intenseÞ:
The intense and sharp O–H stretching frequency of the nonassociated O–H
group n(OH)free is observed in the gaseous state, in very dilute solutions, or in the
case of substances that cannot associate for steric reasons.
Association due to hydrogen bridges gives rise to a decrease of the wave numberand an increase of the intensity of the O–H. . .O stretching vibration.
The influence of the concentration on the O–H stretching vibrations in the range
from 3,650 to 3,000 cm�1 is shown in Fig. 2.21a–c.
Only nonassociated species are present in strongly diluted solutions as can be
concluded from the absence of the broad absorption bands of dimeric and higher
associated species at shorter wave numbers as shown in the infrared spectra of
Fig. 2.21a, b. In contrast only polymeric species are present in highly concentrated
solutions or in liquid samples as Fig. 2.21c shows by missing O–H stretching
frequencies for the monomeric and dimeric species.
The stronger the hydrogen bonding the greater is the shift to lower wave
numbers. For example, a shift up to �2,200 cm�1 is observed for inorganic
compounds with the strongest hydrogen bridges. Thus, the wave number shift is asemiquantitative indicator to evaluate the strength of hydrogen bridges.
The broadness of the infrared bands caused by association through hydrogen
bridges is determined by the degree of association. The broader the infrared bands
the higher the degree of association. Because the hydrogen bridge in N–H
compounds is weaker, only dimeric species are formed and the N–H stretching
vibrations give rise to narrower bands which is evidence in order to distinguish N–H
and O–H stretching vibrations in a given infrared spectrum.
Intramolecular hydrogen bridges are marked by a broad and flat pattern which is
independent from the concentration; see, for example, the O–H stretching band for
o-hydroxy acetophenone (Structure 2.8) in Fig. 2.22a, b. The intermolecular hydro-
gen bridge is caused by equilibrium; therefore, the pattern is determined by the
concentration (see Fig. 2.21a–c).
106 2 Vibrational Spectroscopy
OH
O
CH3
2.8
The strong intramolecular hydrogen bridges of diketones provide very flat andvery broad bands in the range from approximately 3,200 to 2,000 cm�1 which is
demonstrated by the very strong chelate bridge band of acetylacetone (Structure
2.9) in Fig. 2.23.
OH
O
CH3CH3
2.9
3000 4000 30004000
T in %
n in cm-1n in cm-1
a b
Fig. 2.22 (a, b) Infrared
spectrum in the range of O–H
stretching vibration of
o-hydroxy-acetophenone(Structure 2.8) in CCl4: (a)
c ¼ 0.025 mol L�1,l ¼ 0.5 mm; (b)
c ¼ 0.5 mol L�1, l ¼ 0.025
mm
100
50
n (O-H…O)
T in %
1500 10002000250030003500
n in cm-1
Fig. 2.23 Infrared spectrum of acetylacetone (Structure 2.9) containing a very strong chelate
bridge; cuvette: l ¼ 0.01 mm
2.3 Interpretation of Vibrational Spectra of Organic Molecules 107
Bending deviations of the C–O–H group
Wave number range : din�plane: 1375� 45 cm�1 dout�of�plane: 650� 50 cm�1
The two types of bending deviations of the C–O–H group give rise to very broad
absorption bands; thus, they can easily be distinguished by C–H bending modes of
alkyl and aromatic structures. The in-plane bending vibration of primary and
secondary alcohols couples with the C–H wagging mode resulting in two bands
at �1,420 cm�1 and �1,330 cm�1, respectively. Such coupling is not achievable
for tertiary alcohols; therefore, in general, only one band is observed in this range.
This evidence can be used to recognize tertiary alcohols.
The bending vibrations of phenols – din-plane and dout-of-plane – are observed in thesame range as those of the alcohols. The ds(CH3) band, valuable for the recognition
of the presence of a CH3 group, can be distinguished from the din-plane by its greaterbroadness.
“C–O” stretching vibrationsWave number range: n(“C–O”): 1,280–980 cm�1 (very intense bands)
The very intense bands in the range from 1,280 to 980 cm�1 are generally
assigned to the stretching vibration of the C–O group, but these bands are caused
by coupling with the neighboring C–C bond, especially in primary alcohols.
Therefore, these bands should be named more precisely asymmetrical C–C–O
stretching vibration nas(C–C–O). The bands are diagnostically very valuable
because they enable the recognition of the alcohol type because of their various
positions in the infrared spectrum. They are observed in the ranges from 1,075 to
1,000 cm�1 for primary (1�), from 1,150 to 1,075 cm�1 for secondary (2�), andfrom 1,210 to 1,100 cm�1 for tertiary (3�) alcohols, respectively. As one can see,
the wave numbers of secondary and tertiary alcohols partially overlap which
prevents the recognition of the alcoholic type. Sometimes, the symmetricalC–C–O stretching vibration ns(C–C–O) can be helpful which is observed between
900 and 800 cm�1 in 1� and 2� alcohols and lies at �1,000 cm�1 in 3� alcohols,respectively.
The C–O stretching vibration of phenols lies in the range from 1,260 to
1,200 cm�1 which is approximately the same range as for 3� alcohols. However,the distinction between both structures should not be difficult because of the
characteristic pattern of the aromatic moiety.
Let us now turn to the problem of the occurrence of absorption bands of water.Water is contained in many compounds, even if only in traces, or it is incorporated
during the sample preparation. Water provides stretching vibrations that overlap
with those of alcohols. Therefore, the recognition of the O–H group of an alcohol
can be complicated. However, in contrast to alcohols water provides a bending
vibration at �1,650 cm�1 which is demonstrated by the infrared spectrum of a
hydrocarbon which is contaminated with water, shown in Fig. 2.24.
108 2 Vibrational Spectroscopy
2.3.6.2 Ethers, C–O–C(a)
Saturated ethers or such containing a nonbranched a-C atom provide a strongly
asymmetric C–O–C stretching vibration in the range from 1,150 to 1,080 cm�1
(mostly at �1,125 cm�1) and a symmetric C–O–C stretching vibration with
medium intensity which lies in the range from 890 to 820 cm�1. Ethers with a
branched a-C atom show two or more absorption bands with approximately equal
intensity in the range from 1,210 to 1,070 cm�1 because of coupling of vibrations ofthe C–O with the neighboring C–C group.
Note that the range of the stretching vibrations of ethers overlaps with those of
alcohols (and also carboxylic acid esters, see below), but the presence of an alcohol
requires additionally a very intense O–H stretching vibration as explained above
(and esters are characterized by intense C¼O stretching vibrations).
For other types of ethers intense absorption bands are observed in the following
ranges:
Aryl alkyl ether : nasðC�O�CÞ : 1; 275� 1; 200 cm�1; nsðC�O�CÞ :� 1; 210� 10 cm�1;Di-aryl ethers : nasðC�O�CÞ : 1; 300� 1; 200 cm�1:
A representative infrared spectrum employing the example di-sec-butyl ether(Structure 2.10) is shown in Fig. 2.25. The bands assigned to the vibration of the
ether group are marked in bold.
O
2.10
50
3500 3000 1500 1000
968
13821365
1465
3030
2963
2932
2872
d (H2O)n (H2O)100
T in %
n in cm-1
Fig. 2.24 Infrared spectrum of a hydrocarbon compound which is contaminated with water
2.3 Interpretation of Vibrational Spectra of Organic Molecules 109
2.3.6.3 Amines, C–NHR (R ¼ H, Alkyl, Aryl)
N–H stretching vibrations, n(N–H)
Wave number range : 3500� 3280 cm�1
Primary (1�) amines show two IR-absorption bands: nas(N–H) at
3,380–3,350 cm�1 and 3,500–3,420 cm�1 for saturated and aromatic amines,
respectively, and ns(N–H) at lower wave numbers. The intensities of n(N–H) areweaker than those of n(O–H) because of the weaker hydrogen bridges in amines in
contrast to alcohols and phenols. Thus, only dimer species are formed resulting in
narrower bands as is shown in the infrared spectrum of benzyl amine (Structure
2.11) presented in Fig. 2.26.
CH2 NH2
2.11
Secondary (2�) amines give rise to only one N–H stretching vibration which lies
in the range of 3,320–3,280 cm�1 and at �3,400 cm�1 for saturated and aromatic
amines, respectively.
Tertiary (3�) amines do not show any N–H stretching vibration.
Bending vibrationsWave number range : din�planeðNH2Þ : 1650� 1580; doopðNH2Þ : 850� 700 cm�1
50
100
3000 25001500 1000
29692931
2877
14651455
1333
13751118
11661147
875831T in %
n in cm-1
Fig. 2.25 Infrared spectrum
of di-sec-butyl ether(Structure 2.10); sampling
technique: capillary thin film
110 2 Vibrational Spectroscopy
The medium intense scissoring din-plane(NH2) is observed as a relatively broad
band. Thus, it can be distinguished from other vibrations in this range and is
important in order to distinguish primary amines and alcohols.
The out-of-plane bending vibration lies in the range from 850 to 750 cm�1 andfrom 750 to 700 cm�1 for primary and secondary amines, respectively. Sometimes,
this difference may be used to distinguish both types of amines.
C–N stretching vibrations, n(C–N)
Wave number range : 1350� 1020 cm�1
The medium intense n(C–N) lies in the range from 1,250 to 1,020 cm�1 and from1,350 to 1,250 cm�1 for saturated and aromatic amines, respectively. Because of its
lower intensity it can be easily distinguished from the very intense C–O bending
vibration provided it is not superimposed by a n(C–O). Primary amines containing a
branched aC atom show new bands due to coupling of vibrations but they cannot be
reasonably assigned.
The asymmetrical C–N–C stretching vibration of secondary amines can be used
to distinguish saturated and aromatic secondary amines: Saturated secondary
amines show bands at 1,180–1,130 cm�1 and those of aromatic ones are observed
in the range from 1,350 to 1,250 cm�1.
2.3.6.4 Thiols, C–S–H
Wave number range : nðS� HÞ : 2600� 2550 cm�1
30003500 2000 1500 1000 500
50
100
33733290
3025
30853065
29202870
16051585
14951452
1385
865
T in %
n in cm-1
Fig. 2.26 Infrared spectrum of benzyl amine (Structure 2.11); sampling technique: capillary thin
film
2.3 Interpretation of Vibrational Spectra of Organic Molecules 111
The S–H stretching frequency lies in a range which is not occupied by other
fundamentals; hence, the functional group S–H can be well recognized using
infrared spectroscopy. Note that the intensity of n(S–H) is low because of the
small dipole moment change during vibration of this group.
Challenge 2.11
Justify the structure of N-methyl p-toluidine (Structure 2.12) by assignment
of the marked bold infrared absorption bands in Fig. 2.27.
NHCH3 CH3
2.12
Solution to Challenge 2.11
The assignment of the absorption bands of the infrared spectrum in Fig. 2.27
is summarized in Table 2.15.
(continued)
50
3500 2000 1500 1000
3410 3016
29202886
2810
16181585
1524810
13151261
14731446
694
3000
T in %
100
n in cm-1
Fig. 2.27 Infrared spectrum of N-methyl p-toluidine (Structure 2.12); sampling technique:
capillary thin film
112 2 Vibrational Spectroscopy
The following structural moieties can be justified according to results in
Table 2.15:
1. The aromatic structure is recognized by n(sp2C–H) in the range from 3,100 to
3,000 cm�1 in addition to n(C¼C) at�1,600–1,490 cm�1. Note that there are nohints of an alkene because of the missing absorption band n > 1,600 cm�1.
The para-substitution is proposed by g(C–H) at 810 cm�1 in combination
with the missing ring bending vibration. Note that the latter fact is also valid for a
1,2-disubstitution, but the pattern of the weak overtone and combination
vibrations in the range from 2,000 to 1,650 cm�1 indicate the para-substitutionrather than the ortho-one (see Fig. 6.1).
2. The presence of the secondary amine structure is indicated by the intense
single band at 3,410 cm�1 which belongs to n(N–H) and, additionally, by doopat �700 cm�1.
3. Alkyl groups are indicated by n(sp3C–H) in the range from 3,000 to 2,800 cm�1
as well as the asymmetric and symmetric bending vibrations at 1,473, 1,446, and
1,315 cm�1.The stretching vibrations of a CH3 group which is directly linked to aryl or N
are diminished and are observed for the N-aryl structure at 2,925 � 5,
2,865 � 5, and 2,815 � 5 cm�1, which is experimentally justified by the
respective absorption band.
The CH2 group cannot be recognized by its respective vibrations because the
position of its sole indication, namely the presence of nas(C–H) at �2,925 cm�1
(see Table 2.11), is overlaid by the shifted stretching vibrations of the N-CH3
group. Note that recognition number and types of alkyl groups are not the subject
of infrared spectroscopy.
Table 2.15 Assignment of infrared absorption bands of N-methyl p-toluidine(Structure 2.12)
n in cm�1 Assignment n in cm�1 Assignment
3,410 vs n(N–H),2� Amine 1,473 m das(CH3)
1,446 m
3,016 m n(sp2C–H), Aryl 1,315 w ds(CH3)
2,920 m ns(CH3) + Fermi
resonance
1,261 m n(C–N–C), N–aryl2,886 m
2,810 m ns(CH3), N-aryl 810 s g(C–H), 1,4-disubstituted1,618 s–,524 n(C¼C), Aryl 694 m doop 2� Amine
2,000–1,650 w Overtones and combination
vibrations for aryl
1,4-disubstituted
2.3 Interpretation of Vibrational Spectra of Organic Molecules 113
Challenge 2.12
For solutions to Challenge 2.12, see extras.springer.com.
1. Assign the given absorption bands of the infrared spectrum of di-sec-butylether (Structure 2.10) in Fig. 2.25. The wave numbers that are important
for the ether group are marked in bold.
2. Assign the given absorption bands of the infrared spectrum of
benzylamine (Structure 2.11) in Fig. 2.26. The wave numbers that are
important for the amine group are marked in bold.
3. An unknown compound with the sum formula C4H11N give rise to the
infrared spectrum shown in Fig. 2.28. Derive the structure.
4. The infrared spectra of three structurally isomeric compounds are
presented in Fig. 2.29a–c. Derive the structure of each compound. The
sum formula can be determined by the spectrometric data of the isomer B;
m/z (Irel in%): 108 (83.5) M•+, 109 (6.8).
(continued)
3000
100
50
3500 2950 1500 1000
33683282
2962 2875
1608
1481
1381
1337
11721133
805
T in %
n in cm-1
×
Fig. 2.28 Infrared spectrum of an unknown liquid compound; sampling technique:
capillary thin film; � - Fermi resonance: Overtone of d(NH2) + n(NH)
114 2 Vibrational Spectroscopy
3000 1500 1000
100
100
50
3094306330333003
29452836
16001587
19301840
17451684
1497 12471040
755692
146814541442
a
T in %
n in cm-1
50
3500 3000 2500 2000
3325
30883065 3031
29322875
19621876
1811
1755
16061592
14971453
1039736 696
b
1500 1000
T in %
n in cm-1
c
3500 3000 1500 1000
3295
3022
2940(sh)2922
2866
1876
16131599
1513
13561469 1435
8141222
100
50
T in %
n in cm-1
Fig. 2.29 (a–c) Infrared spectra of unknown liquid compounds; sampling technique:
capillary thin film
2.3 Interpretation of Vibrational Spectra of Organic Molecules 115
2.3.7 Interpretation of Vibrational Spectra of CarbonylCompounds Containing the Functional Group –(C¼O)–Xwith X ¼ H, C, OH, OR, O(C¼O)R, NR2, and Cl
Wave number range : nðC ¼ OÞ : 1; 850� 1; 650 cm�1
As explained above, the C¼O stretching mode is a very characteristic vibration.
However, its range of 200 cm�1 is relatively large for a characteristic vibration.
Electronic effects are the main reasons for this, as explained in Sect. 2.3.2. Knowl-
edge of these relations, in combination with further information, is the basis for
recognition of the substituent X linked at the C¼O group which is summarized
in this section. Note that a weak band mostly occurs in the range n �3,400–3,650 cm�1 which is the overtone of the intense C¼O stretching vibration.
Although this overtone coincides with the O–H stretching vibration it can be
unambiguously assigned by its low intensity and the missing broadness.
2.3.7.1 Ketones, X ¼ C
Wave number ranges for diagnostically important vibrations of ketones are listed in
Table 2.16.
Decreasing of n(C¼O) is observed by branching at the aC atom.
Ketones can be recognized by the combination of n(C¼O) and n(C–C–C). Incontrast to n(C–C) of the alkyl chain the intense asymmetrical stretching vibration
of the C–C(¼O)–C skeleton indicates a ketone provided that functional groups of
the aldehyde, ester, and carbonic acids are not present in the molecule. Figure 2.30
shows the IR-spectrum of acetophenone (Structure 2.13). The bold IR-bands belong
to the vibration of the carbonyl group.
CH3
O
2.13
Table 2.16 Wave number ranges of important vibrations of ketones in cm–1
Saturated ketones Aryl-alkyl-ketones Di-aryl ketones
n(C¼O) 1,715 � 10 1,700�1,670 1,680�1,650n(C–C–C) 1,230–1,100 1,300�1,230 1,300�1,230
116 2 Vibrational Spectroscopy
2.3.7.2 Aldehydes, X ¼ H
Wave number ranges for diagnostically important vibrations of aldehydes are listed
in Table 2.17.
Although the range of the very intense C¼O stretching vibration lies admittedly
somewhat higher, it overlaps partially with that of the corresponding ketones. Thus,
this band is not an evidence to distinguish both functional groups. However, thealdehyde can be unambiguously distinguished from the corresponding ketone bytwo or one intense absorption bands at 2,850� 2,700 cm�1 combined with d(CHO)at �1,390 � 10 cm�1. This is shown in the infrared spectrum of n-hexanal(Structure 2.14) presented in Fig. 2.31. Note that only the band of the Fermi
resonance is well separated, whereas n(CH) is developed only as a shoulder becauseof overlap with the CH-stretching vibration. This fact is also to be expected for a
saturated, non a-C branched aldehyde. In the case of a-branched saturated
aldehydes, however, n(CH) is decreased (see Table 2.17) and hence, two separate
bands will be observed in this range.
100
3355
3067 -3005
2967 2866T in %
50
1686
15991583 1267
76113601
14501430
691
3000 2000 1500 1000
n in cm-1
Fig. 2.30 Infrared spectrum of acetophenone (Structure 2.13); sampling technique: capillary thin
film
Table 2.17 Wave number ranges of important vibrations of aldehydes in cm�1
n(CH) + Fermi resonance 2,850�2,700, 1 or 2 bands
n(CH), saturated, non a-C branched 2,730�2,715n(CH), saturated, a-C branched 2,715�2,700n(C¼O), saturated 1,730 � 10
n(C¼O), aryl 1,700 � 10
d(CHO) �1,390
2.3 Interpretation of Vibrational Spectra of Organic Molecules 117
2.3.7.3 Carboxylic acids, X ¼ OH
Since carboxylic acid contains a carbonyl and a hydroxyl group, bands of both
functional groups can be seen in the corresponding spectra. A representative
spectrum is shown in Fig. 2.32 for propionic acid.
Typical bands for the carboxylic acid groups are marked in bold.
O�H stretching vibrationsThe very broad and intense infrared band in the range from 3,500 to 2,500 cm�1
is an unambiguous evidence for the carboxylic group. No other functional group
shows such a broad and intense band, which is caused by the strong hydrogen
bonding. Even in the gas state dimeric species linked by hydrogen bridges are still
present. Additionally, overtone and combination modes feature bands of medium
intensity on the low side of the O�H stretching up to 2,500 cm�1. As shown in
Fig. 2.32, the sharper C�H stretching bands are superimposed upon the broad O�Hstretching band. Note that the free n(OH) stretching mode appears only in very
strongly diluted solutions at �3,520 cm�1. The broad O�H stretching bands of
b-diketones caused by the strong chelate hydrogen bridge lie in the same range as
those of the carboxylic group, however, the intensity is markedly lower (compare
the infrared spectra of Figs. 2.23 and 2.32).
C¼O stretching vibrationsThe intensity of the C¼O stretching n(C¼O) is higher than that of ketones,
compare Figs. 2.30 and 2.32. Monomeric saturated carboxylic acids absorb
at 1,760 cm�1. The common dimeric structures feature only the asymmetrical
C¼O stretching vibration, which lies at 1,710 � 10 cm�1 for aliphatic carboxylicacids. Conjugation diminishes the band position of n(C¼O) which is found at
1,700 � 15 cm�1. Intramolecular hydrogen bonding can be recognized by bands
100
7323432T in %
50
29602942
27192822
28742864 1719
1468 1391
13601342
10001500250030003500
n in cm-1
Fig. 2.31 Infrared spectrum of n-hexanal, CH3�(CH2)4�COH (Structure 2.14); sampling tech-
nique: capillary thin film
118 2 Vibrational Spectroscopy
markedly shifted to lower wave numbers. Thus, the C¼O stretching bands of
o-hydroxy benzoic acid and p-hydroxy benzoic acid are observed in KBr at 1,667
and 1,682 cm�1, respectively.C�O and O�H bending vibrationsCoupling of the C�O stretching vibration n(C�O) with the in-plane bending
din-plane(C�O�H) provides two bands. The bending named by “din-plane(O�H)” liesin the range from 1,440 to 1,395 cm�1 and the more intense “n(C�O)” lies at
1,315–1,280 cm�1. The latter is split into more bands for long-chained carboxylic
acids which is an evidence to recognize fatty acids.Carboxylic acid salts (carboxylates)Infrared spectra of carboxylates differ from the carboxylic acids by the lack of
broad stretching and bending bands. The carboxylic group provides two bands in
the range from 1,650 to 1,350 cm�1. The more intense band at the higher wave
number is assigned to the asymmetric CO2 stretching vibration and that in the range
from 1,450 to 1,350 cm�1 to the symmetric CO2 stretching vibration, respectively.
2.3.7.4 Carboxylic Acid Esters, X ¼ OR
Aliphatic and aromatic carboxylic acid esters are characterized by three intenseinfrared bands which lie in the ranges n � 1,700 cm�1, � 1,200 cm�1, and �1,100 cm�1. This so-called “three band rule” is diagnostic for the carboxylic acidester group.
The band with the highest wave number belongs to the C¼O stretching vibration
n (C¼O). The medium band is caused by asymmetric C�C and C�O attached to
the carbonyl carbon and is called nas(C�C�O). The third of these bands includes
the asymmetrical vibration of the ester oxygen and the next two carbon atoms in the
100
50
T in % Δν1 / 2 ≈ 900 cm-1
3500 3000 2500 2000 1500 1000
29862948
2660
1716
9331080
1240
14191467
1384
n in cm-1
Fig. 2.32 IR-spectrum of propionic acid CH3�CH2�COOH (Structure 2.15); sampling tech-
nique: capillary thin film
2.3 Interpretation of Vibrational Spectra of Organic Molecules 119
hydrocarbon chain nas(O�C�C). All of these bands can be found in the infrared
spectrum of propionic acid ethyl ester C2H5�COO�C2H5 (Structure 2.16) shown
in Fig. 2.33. The bands typical for the carboxylic acid groups are marked in bold.
Wave number ranges for various types of esters are summarized in Table 2.18.
2.3.7.5 Carboxylic Acid Anhydrides, X¼O–(C¼O)–OR
C¼O stretching vibrationsThe coupling of the two C¼O groups results in two intense C¼O stretching
vibrations of which the symmetrical vibration is observed at higher wave numbers
than the asymmetrical one. The symmetrical stretching vibration provides more
intense bands than the asymmetric one for noncyclic anhydrides; the reverse
intensity ratio is true for cyclic anhydrides.
100
3665
T in %
50
2985
2945 1465
13721347
1047
3000 1500 1000
1192
1740
n in cm-1
Fig. 2.33 Infrared spectrum of propionic acid ethyl ester (Structure 2.16); sampling technique:
capillary thin film
Table 2.18 Wave number ranges for various types of carboxylic esters
Saturated esters Aromatic or a,b-nonsaturated esters
n(C¼O) 1,750–1,735 1,730–1,710
n(C�C�O) 1,210–1,165 1,330–1,250
n(C�C�O), acetates �1,240n(O�C�C) 1,100–1,030 1,130–1,000
120 2 Vibrational Spectroscopy
The intensity ratio of ns(C¼O) and nas(C¼O) enables us to distinguish betweencyclic and noncyclic anhydrides, respectively.
In the infrared spectrum of noncyclic propionic acid anhydride (Structure 2.17)
shown in Fig. 2.34 the stretching vibrations lie at 1,820 cm�1 [the more intense
ns(C¼O)] and 1,762 cm�1 [the weaker nas(C¼O)]. The respective C¼O stretching
vibrations in the infrared spectrum of cyclic 3-methyl glutaric acid anhydride
(Structure 2.18) presented in Fig. 2.35 at ns(C¼O) ¼ 1,809 cm�1 and nas(C¼O)¼ 1,762 cm�1 have the reverse intensity ratio. The same result is true for five-ring
anhydrides.
O
O O
2.17
The infrared bands due to the anhydride group are marked in bold.
O
O
O 2.18
100
T in %
50
2969
2948
1464
14141348
3000 2000 1500 1000
2889 1762 10421820
n in cm-1
Fig. 2.34 Infrared spectrum of noncyclic propionic acid anhydride (Structure 2.17); sampling
technique: capillary thin film
2.3 Interpretation of Vibrational Spectra of Organic Molecules 121
C–O stretching vibrationsThe C–O and C–C–O stretching vibrations, in most cases more intense than
n(C¼O), lie at n � 1,050 cm�1.The wave number ranges of the C¼O and C–O stretching vibrations are listed in
Table 2.19 for some anhydride types.
2.3.7.6 Carboxylic Acid Amides, X ¼ NR2
N–H stretching vibrationsThe number of N–H stretching vibrations is equal to the number of N–H bonds.
The two bands of primary (1�) amides R–(C¼O)–NH2 are seen atn � 3,520 and
�3,400 cm�1 with medium intensity in the spectrum which is obtained by
employing diluted solutions. In the solid state the bands are shifted to smaller
wave numbers because of hydrogen bonding.
100
50
T in %
15003500 10003000
18091762 1066
974
936
n in cm-1
Fig. 2.35 Infrared spectrum of 3-methyl glutaric acid anhydride (Structure 2.18); sampling
technique: capillary thin film
Table 2.19 Wave number ranges of the C¼O and C–O stretching vibrations for some anhydride
types
Noncyclic anhydrides Cyclic anhydrides
ns(C¼O) Saturated 1,820 � 5 (stronger) 1,860 � 10 (weaker)
Nonsaturated 1,775 � 5 (stronger) 1,850 � 10 (weaker)
nas(C¼O) Saturated 1,750 � 5 (weaker) 1,785 � 15 (stronger)
Nonsaturated 1,720 � 5 (weaker) 1,870 � 10 (stronger)
n(C–O) 1,050 � 10 1,300–1,175 and 950–880
122 2 Vibrational Spectroscopy
Secondary (2�) amides R–(C¼O)–NRH show only one band in the range from
3,500 to 3,400 cm�1 in diluted solutions. In the solid state generally several bands
appear caused by various conformeric structures that are formed by hydrogen
bonding.
Tertiary (3�) amides R–(C¼O)–NR2 do not show any bands in the range from
3,500 to 3,300 cm�1.C¼O stretching vibrationsn(C¼O) is shifted to smaller wave numbers due to mesomerism as is shown by
the two resonance structures:
CR
O
NR2
R C
O-
NR2+
Furthermore, the C¼O stretching vibration cannot be assumed to be purely
n(C¼O) because of coupling with other vibrations. Therefore, “n(C¼O)” is
named the amide I band. It is seen in the spectrum always below 1,700 cm�1 andits position is determined by the sampling technique and the amide type. The amide
I band of tertiary amides is nearly independent of the physical state because of
missing hydrogen bridges.
N–H bending vibrationsThe NH2 scissoring d(NH2) couples with the N–C¼O stretching vibration,
because both vibrations lie in nearly the same wave number range, result in a
band which is named amide II. In 1� amides the amide II bands mostly coincides
with the amide I band or it is developed only as a shoulder but in solution both
bands are separate.
Coupling of the amide II band with the d(NH2) band gives rise to the amide IIIband which appears at n � 1,250 cm�1.
Other vibrations1� amides show a band at n � 1,400 cm�1 which is assigned to the C–N
stretching vibration.
The out-of-plane bending of the NH2 group doop(NH2), also named “wagging
NH2,” appears as a broad band in the range from 750 to 600 cm�1 whose intensity issmaller for 2� amides. Typical wave numbers of various types of amides are
summarized in Table 2.20.
Figure 2.36 presents the infrared spectrum of benzamide (Structure 2.19) as an
example of a primary amide prepared in KBr. Typical wave numbers for the aryl-
benzamides are marked in bold.
NH2
O
2.19
2.3 Interpretation of Vibrational Spectra of Organic Molecules 123
2.3.7.7 Carboxylic Acid Chlorides, X ¼ Cl
Recall from the discussion in Sect. 2.3.1 that a substituent with an –I effect linked at
the carbonyl group decreases the wave number of n(C¼O) bands. Because of the
strong –I effect of the Cl atom, the shift of n(C¼O) towards higher wave numbers is
large. Thus, n(C¼O) of nonconjugated carboxylic acid chlorides appears in the highwave number range from 1,815 to 1,785 cm�1. Conjugation, as discussed above,
diminishes the n(C¼O) band position to 1,800–1,770 cm�1.
100
3064
3032
T in %
50
3366
3170
1656 16241577
1460
1405
3500 100020003000 1500
n in cm-1
Fig. 2.36 Infrared spectrum of benzamide (Structure 2.19) as an example of a primary amide;
sampling technique: KBr pellet
Table 2.20 Typical wave numbers (in cm�1) of various types of amides
1� amide 2� amide 3� amide
n(NH)nonassociated ns: 3,520 3,500–3,400 No
nas: 3,400n(NH)associated ns: 3,350 3,300–3,060 No
nas: 3,180 Mostly several
Amide Inonassociated 1,690 1,700–1,680 1,680–1,630
Amide Iassociated 1,680–1,630 1,680–1,630 No
Amide II 1,655–1,620 1,570–1,515 No
Separated in diluted solutions High intensity
n(C–N) 1,430–1,390 1,310–1,230 �1,505R¼CH3
doop(O¼C–NH2) 750–600 (broad) 750–680 broad No
124 2 Vibrational Spectroscopy
The (C¼O) stretching vibration of aryl-carboxylic acid chlorides is split into
two absorption bands due to Fermi resonance as discussed in Sect. 2.1.4. This is an
evidence to distinguish alkyl and aryl carboxylic acid chlorides. In general, the
band at lower wave number has a lower intensity.
The infrared spectrum of benzoyl chloride (Structure 2.20) is shown in Fig. 2.37.
The typical wave numbers for carboxylic acid chlorides are marked in bold.
Cl
O
2.20
2.3.8 Interpretation of Vibrational Spectra of N-ContainingCompounds
2.3.8.1 NO2 Compounds
The NO2 group gives rise to two intense infrared bands that are easy to identify and
this is demonstrated in the infrared spectrum of p-ethyl nitro benzene (Structure
2.21, R¼C2H5) shown in Fig. 2.38. The group wave numbers of the NO2 group are
marked in bold in the spectrum.
NO2
2.21
R
100
50
3493
3071T in %
1775 1733
15961583
1450 873
3500 2000 1500 10003000
n in cm-1
Fig. 2.37 Infrared spectrum of benzoyl chloride (Structure 2.20); sampling technique: capillary
thin film using AgCl discs
2.3 Interpretation of Vibrational Spectra of Organic Molecules 125
The most intense asymmetrical N–O stretching vibration nas(NO2) appears at
higher wave numbers than the less intense symmetrical one ns(NO2). Conjugation
in aryl nitro compounds shifts the wave number to somewhat smaller values, see
Table 2.21.
Electron donor substituents decrease the group wave numbers and increase their
intensity. Thus, nas(NO2) and ns(NO2) of p-nitro aniline (Structure 2.21, R¼NH2)
appear at 1,475 and 1,310 cm�1, respectively as very intense infrared bands.
Coupling between doop(NO2) and g(C–H) of the aromatic group gives rise to new
bands which alter the pattern of the infrared bands in the range from 2,000 to
1,650 cm�1. Hence, the pattern cannot be used to recognize the aromatic substitu-
tion type.
2.3.8.2 N¼O Compounds
Because of similar mass and bond order the N¼O stretching vibration should give
rise to infrared bands similar to those of the C¼O group. Thus, n(N¼O) is observed
100
50
T in %
30793055
297129372877
16061600
1517
14951460
1346
1317
852
1000150020003000
n in cm-1
Fig. 2.38 Infrared spectrum of p-ethyl nitro benzene (Structure 2.21); sampling technique:
capillary thin film
Table 2.21 Group wave numbers of the functional group NO2 (in cm�1)nas(NO2) ns(NO2)
Alkyl-NO2 1,570–1,540 (very intense) 1,390–1,340 (intense)
Aryl-NO2 1,560–1,500 (very intense) 1,360–1,300 (intense)
126 2 Vibrational Spectroscopy
as an intense infrared band at 1,585–1,540 and 1,510–1,490 cm�1 for alkyl–N¼Oand aryl–N¼O compounds, respectively.
2.3.8.3 Nitriles
The C�N group gives rise to characteristic vibrations in a unique wave number
range. The C�N stretching vibration is observed from 2,260 to 2,240 cm�1 for
saturated nitriles and from 2,250 to 2,200 cm�1 for aromatic ones, respectively.
Thus, the position of n(C�N) enables us to distinguish between saturated and
aromatic nitriles.
2.3.9 Interpretation of Vibrational Spectra of S–O-ContainingCompounds
2.3.9.1 Structure Types RR0S¼O, RO–(S¼O)–O–R0
Because of the larger mass of S the bands of the S¼O stretching vibrations are
shifted to lower wave numbers compared to the respective C¼O ones and appear in
the range from 1,225 to 980 cm�1 as intense infrared bands. For further informa-
tion, see Sect. XIII in Table 6.15.
2.3.9.2 Structure Types –SO2–
Like nitro compounds the SO2 group gives rise to an asymmetric nas(SO2) and a
symmetric ns(SO2) stretching vibration which can be well recognized in the infrared
spectrum in the range from 1,420 to 1,000 cm�1 because of their high intensity.
nas(SO2) lies at a higher wave number than ns(SO2). Note that the range of the group
wave numbers of the SO2 compounds overlap with those of other functional groups.
Challenge 2.13
Mass and infrared spectra are given for an unknown compound in Figs. 2.39
and 2.40, respectively. Create a proposal for the structure of this compound as
far as it is possible. The peaks and wave numbers given in the mass and
infrared spectra, respectively have to be assigned.
(continued)
2.3 Interpretation of Vibrational Spectra of Organic Molecules 127
100
50
3500 2500 2000 1500 1000
3490
3395
3083
293329222838
16301597
1508
1438
1347
873
92011381032
738
815
3000
T in %
n in cm-1
Fig. 2.40 Infrared spectrum of an unknown compound; sampling technique: KBr pellet
50 57 64 71 78 85 92 99 106 113 120 127 134 141 148m/z
0
20
40
60
80
100
Irel
152
77
79
51
106
122
104
135
Fig. 2.39 70-eV mass spectrum of an unknown compound. m/z (Irel in%): 152 (100),
153 (8.8)
128 2 Vibrational Spectroscopy
Solution to Challenge 2.13
Mass spectrum
• Derivation of the sum formulaThe results for the derivation of the sum formula are summarized in
Table 2.22.
• Calculation and assignment of the double bond equivalents (DBE)Transformation of the sum formula into CnHx according to the rules given
in Sect. 1.3:
C7H8N2O2 (-2 O–2 N + 2 CH)) C9H10
DBE ¼ 5 calculated by Eq. 1.19.
Assignment: Aryl (4 DBE) + NO2 (1 DBE),
) No further rings can be present in the molecule.
• Structural moieties determined by mass spectroscopy
Structural moiety/functional group Recognized by
NO2, NH2 Molpeak (N rule), (M � X) fragment peaks
Aromatic compound DBE, specific fragment peaks: m/z ¼ 51,
65, 77 amu
CH3 Difference to the sum formula
(continued)
Table 2.22 Results for the development of the sum formula. The m/z values and differences are
given in amu
Number of C and N atoms nC, nNm/z (Irel in%) Assignment Information
152 (100) M•+ nN ¼ 0 or 2
153 (8.8) [M + 1]•+ nC ¼ 8 � 1
Evidence of N in the molecule
Search for N-containing fragment peaks
m/z (start ion) m/z (end ion) Difference Loss of Hints for
152 122 30 N¼O Aryl–NO2
152 106 46 NO2 Aryl–NO2
106 79 27 HCN Aryl–NH2 or pyridine
104 77 27 HCN Aryl–NH2 or pyridine
Result nN ¼ 2
Search for O-containing fragment peaks
m/z (start ion) m/z (end ion) Difference Loss of Hints for
152 136 16 O Aryl–NO2
152 135 17 OH Ortho-effect?
Development of the sum formula
Summation of the atoms recognized as yet 8 � 1 C + 2 N + 2 O
Proposals Mass number Difference to M nHC7N2O2 + Hx 144 + xH 152–144 ¼ 8 8
C8N2O2 + Hy 156 Not possible, because larger than M
Result C7H8N2O2 OE•+ proved according to Eq. 1.11
2.3 Interpretation of Vibrational Spectra of Organic Molecules 129
• The compound is a nitro amino toluene (Structure 2.22) whose substitution
type is unknown. Remember that, in general, the recognition of aromatic
isomers is not the subject of mass spectroscopy.
NO2
CH3
NH2
2.22
Infrared spectrum
The assignment of the infrared bands is listed in Table 2.23.
Structural moieties and functional groups determined by infrared
spectroscopy:
Aryl–NO2, R–NH2, Aryl–CH3, 1,2,4-trisubstituted (?) aromatic
compound.
Proposal for structures
The benzene ring must be trisubstituted, the substituents are NO2, NH2,
and CH3. Ortho-substitution of NO2 and CH3 should be excluded because of
the missing [M � 17] peak due to the elimination of O–H via the ortho-effectin the mass spectrum. Thus, Structures 2.23–2.25 come into consideration.
Note that the true structure can only be established by comparison of the
(continued)
Table 2.23 The assignment of the infrared bands obtained by the infrared spectrum shown
in Fig. 2.40
n in cm�1 I Assignment Hints for structural moieties
3,490 vs nas(NH)3,395 vs ns(NH) 1� amine, R–NH2
3,180 w OT das(NH2) + n(NH) Fermi resonance
3,083 w n(sp2CH) Benzene ring/alkene
2,933 w nas(Aryl-CH3)
2,922 w ns(Aryl-CH3) Aryl–CH3
2,861 w OT das(CH3) + n(CH) Fermi resonance
1,950–1,700 vw OT + CV Pattern for 1,2,4-trisubstitution (?)
1,630 s d(NH2) 1� amine, NH2
1,597 w n(C¼C) aryl Aromatic compound
1,508 vs nas(NO2)
1,347 s ns(NO2) Aryl–NO2
1,438 m das(CH3) CH3
1,371 m ds(CH3)
1,032 m n(C–N) ?873 m d(NO2)
920 m g(Aryl-CH) Aryl, 1 neighboring H atom?
815 m g(Aryl-CH) Aryl, 2 neighboring H atoms?
OT overtone, CW combination vibration, I estimated intensity
130 2 Vibrational Spectroscopy
infrared spectrum with those of the respective reference spectra. But this is
not the subject of this book. With the employment of further methods presented
in the next chapters the evidence of the final structure will be adduced.
CH3
NH2
NO2
2.23
NO2
NH2
CH3
2.24
NO2
NH2
2.25
CH3
Challenge 2.14
For solutions to Challenge 2.14, see extras.springer.com.
1. Assign the given wave numbers in the infrared spectra of Figs. 2.30, 2.31,
2.32, 2.33, 2.34, 2.35, 2.36, 2.37 and 2.38 to the respective vibrations. The
wave numbers distinctive for a certain substance class are marked in bold.
Assemble certain and uncertain structural moieties.
2. Carbonyl compounds can be easily recognized by the intense infrared band
in the range from 1,800 to 1,650 cm�1 but there are many substance
classes containing the C¼O group. Show how the following substance
(continued)
2.3 Interpretation of Vibrational Spectra of Organic Molecules 131
classes can be evidenced and distinguished from the others by infrared
spectroscopy: aldehydes, ketones, carboxylic acids, carboxylic acid esters,
amides, and anhydrides.
3. Show how the following structures can be distinguished:
(a) Intra- and intermolecular hydrogen bonding
(b) E/Z isomers for compounds with the general structure RHC¼CHR(c) Substitution types of aromatic compounds
(d) Distinction between aryl–CH2–CHO and CH3–aryl–CHO
(e) Distinction between primary and secondary amines
(f) Recognition of a tertiary carboxylic acid amide
(g) Distinction between CH3–aryl–CHO and aryl–(C¼O)–CH3
(h) Distinction between CH3–aryl–(C¼O)–Cl and Cl–CH2–aryl–CHO
(i) Distinction between R–S–H, R–S–R´, and R–S–S–R´.
4. The oxidation of 3-methoxy-tetraline (Structure 2.26) gives rise to the
main product whose molecular peak is 16 amu greater than that of the
starting compound. The infrared spectrum of a solution of the reaction
product in CCl4 shows an intense band at 1,686 cm�1. What structures can
be assigned to the reaction product and what can be excluded?
O
2.26
5. The following wave numbers for the C¼O stretching vibrations are
obtained in the infrared spectrum of compounds of the structural type
shown in Structure 2.27 with R ¼ H, OH, N(CH3)2, NO2: 1,737, 1,727,
1,715, and 1,685 cm�1. Assign the infrared bands to the respective
carbonyl compounds and justify your decision.
OCH3O
R
2.27
6. An intense absorption band is observed at 1,766 cm�1 in the infrared
spectrum. Decide whether the infrared spectrum belongs to a compound
of Structures 2.28 or 2.29.
(continued)
132 2 Vibrational Spectroscopy
O
OH
CH3
2.28
O
CH3
O
2.29
7. The range of the characteristic C¼O stretching vibrations from 1,850 to
1,650 cm�1 is smaller than that observed for the C¼S stretching vibration
(1,275–1,030 cm�1). Explain this fact.
8. Figure 2.41 shows the infrared spectrum of a liquid compound. Develop
the structure and assign the infrared bands. The molecular peak is
m/z ¼ 128 amu.
9. Figure 2.42 shows the infrared spectrum of a liquid compound. Develop
the structure and assign the infrared bands. The molecular peak is
m/z ¼ 78 amu.
10. Figure 2.43 shows the infrared spectrum of a solid compound. Derive
the structure and assign the infrared bands. The following information is
obtained by the mass spectrum given in m/z (Irel in%): 157 amu (40)
M•+, 158 amu (3.3), 159 amu (2).
11. The infrared spectrum of a di-tert-butylphenol (Structure 2.30) is
presented in Fig. 2.44. Develop the structure and assign the infrared
bands.
OH
(CH3)3C C(CH3)32.30
(continued)
2.3 Interpretation of Vibrational Spectra of Organic Molecules 133
12. Derive the structure and assign the given wave numbers in the infrared
spectrum shown in Fig. 2.44. The CI-mass spectrum (reaction gas:
iso-butylene) shows an intense peak at m/z ¼ 85 amu.
13. Dependent on the solvents the thermic decomposition of phenyl azo
cyclo hexene provides the two products whose structure is given in
Structure 2.31A, B. The infrared spectra feature the following bands in
the range n > 3,000 cm�1:
(continued)
100
50
3500 3000 1500
1380
1000 500
33673081
29692861
1844
1645
1469992
922
680
2933
1342
1322
1022
T in %
n in cm-1
Fig. 2.41 Infrared spectrum of a liquid compound; sampling technique: capillary thin film;
MS: m/z ¼ 128 amu M•+
100
3500
50
3000 2500 1500 1000 500
3364
29352875
2558
14641417
1051
663
T in %
n in cm-1
Fig. 2.42 Infrared spectrum of a liquid compound; sampling technique: capillary thin film;
MS: m/z ¼ 78 amu M•+
134 2 Vibrational Spectroscopy
Product I: sharp band at 3,360 cm�1,Product II: broad band centered around 3,150 cm�1.What structures have the reaction products?
(continued)
100
3500
50
3000 2000 1500 1000
33503258
3070 1555
1481
1447
1332 1152 757690
T in %
n in cm-1
Fig. 2.43 Infrared spectrum of a solid compound; sampling technique: KBr pellet; MS:
m/z (Irel in%): 157 amu (40) M•+, 158 amu (3.3), 159 amu (2)
100
150030003500 1000
50
3646
3083
2969
29142874
1584
1420
1391137313611250 746
T in %
n in cm-1
Fig. 2.44 Infrared spectrum of a di-tert-butylphenol (Structure 2.30); sampling technique:
KBr pellet
2.3 Interpretation of Vibrational Spectra of Organic Molecules 135
N
N
N
H
C6H5
NNH
C6H5
N
H
H
a b
14. Some mass spectrometric data and the infrared spectra of two isomeric
liquid compounds are featured in Fig. 2.45a, b. Develop the structures
and assign the given wave numbers in the infrared spectra.
15. The following bands listed in arbitrary order are observed in the infrared
spectra of four isomeric pentenes possessing the structures
H2C = CHC3H7ðStructure 2:32Þ H2C ¼ C(CH3ÞC2H5ðStructure 2:33Þ
trans� C2H5CH ¼ CHCH3(Structure 2:34Þcis� C2H5CH ¼ CHCH3ðStructure 2:35Þ
A: 966 (s); B: 1,780 (w), 887 (s); C: 1,826 (w), 1,643 (m), 993 (s),
912 (s);
D: 1,659 (m), 689 (m).
Which structure belongs to which infrared spectrum? Assign the infrared
bands to the respective vibrations.
16. Figure 2.46a, b features the Raman (A) and the infrared spectra (B) of an
aromatic C14 compound. Develop the structure and assign the given wave
numbers in the spectra.
17. Figure 2.47a, b presents the infrared spectra of two C8 compounds.
Derive the structure and assign the given wave numbers in the spectra.
18. Some mass spectrometric data and the infrared spectra of four structur-
ally isomeric compounds are given in Fig. 2.48a–d. Derive the structure,
assign the given wave numbers in the spectra and formulate the relevant
mass spectrometric fragmentation for each compound.
19. A section of an infrared spectrum in the range 1,800–1,600 cm�1 obtainedby a reaction product of succinic acid is shown in Fig. 2.49. Which
reaction product was obtained?
(continued)
136 2 Vibrational Spectroscopy
20. In general, the asymmetric C–O stretching vibration is observed at higher
wave numbers than the corresponding symmetrical one. This order is
reversed in carboxylic anhydrides. How can this fact be confirmed
experimentally?
(continued)
100
3016
a
T in %
50
3411
3016
298128102796
1619
1524
14871473
132613051261 809
3500 3000 1500 1500
n in cm-1
3308 2969
2933
1605
b
3500 100020003000 1500
3085
3068
2882
2844
2790 14951474
1454
1356
1260
1206736
6983028
100
50
T in %
n in cm-1
Fig. 2.45 (a, b) Infrared spectra of a two isomeric compounds; sampling technique:
capillary thin film a: h-Ms: [M]•+: m/z ¼ 121.0891 amu; base peak: m/z ¼ 120 amu.
b: MS: Base peak: m/z ¼ 120 amu; intense peaks: m/z (Irel in%): 44 (93), 91 (68)
2.3 Interpretation of Vibrational Spectra of Organic Molecules 137
21. Figure 2.50a–e feature the infrared spectra of unknown compounds
supplemented by some mass spectrometric information. Derive the struc-
ture and assign the given wave numbers in the spectra.
100b
2225
16961495
Irel
a
10002000
n in cm-1
50
3084306530373023
T in %
3000 2500 2000 1500 1000
1500
1444
n in cm-1
Fig. 2.46 Raman (a) and infrared spectrum (b) of an aromatic C14 compound; sampling
technique (IR): capillary thin film
138 2 Vibrational Spectroscopy
100
50
3385
306630303020
292328222728
1480
1380
a
T in %
20003000 100015003500
170416061588
783686
n in cm-1
100
502930
1602 1388
b
T in %
308830643030
28252728
1724
1585
14981454 751
701
1031
3000 2000 1500 1000
n in cm-1
Fig. 2.47 (a, b) Infrared spectra of two C8 compounds; sampling technique: capillary thin film.
a: MS: intense [M � 1]•+ peak. b: MS: Base peak: m/z ¼ 91 amu
2.3 Interpretation of Vibrational Spectra of Organic Molecules 139
100
3500
50
3000 2000 1500 1000
3364
308630633030
297329282878
1601
14931451 1078
761699
1365
T in %
100
3500
50
3000 2000 1500 1000
3420
3067
3036
2968
2934
1592
1503
1493 1466
1329
1232762
2874
1609
T in %
n in cm-1
n in cm-1
a
b
100
3000
50
2500 2000 1500 1000
3097
3066
3044
2984
2929
2900
2878
1489
1477
1389
1245
1051
1456T in %
n in cm-1
c
Fig. 2.48 (continued)
140 2 Vibrational Spectroscopy
100
3500
50
3000 2000 10001500
3340
3085
3064
3028
2944
2877
1604
1584
14971046 748
699
T in %
n in cm-1
d
Fig. 2.48 (a–d) Infrared spectra of four structurally isomeric compounds; sampling technique for
all samples: capillary thin film. a: MS: m/z in amu (Irel in%): 107 (100), 122 M•+, intense peaks:
51, 77, 79. b: MS: m/z (Irel in%): 107 (100), 122 (40) M•+, peaks with Irel > 10%: 39, 51, 77, 79,
91. c:MS: Base peak:m/z ¼ 94 amu; peaks with Irel > 10%:m/z in amu ¼ 39, 51, 65. d:MS: Base
peak: m/z ¼ 91 amu
100
186250
T in %
1787
1800
n in cm-1
Fig. 2.49 Part of the infrared
spectra of a reaction product
of succinic acid; sampling
technique: solution in carbon
tetrachloride
2.3 Interpretation of Vibrational Spectra of Organic Molecules 141
100
50
3088 3067
3030 1577
T in %
a
3005
1666
1595
1449
705 697
3000 2000 1500 1000
n in cm-1
100 3410
b
T in %
2989 2941
2877
50
3243
21201764
1479 1446
13881373
1251
1017
634
3500 3000 2500 2000 1500 1000
n in cm-1
100
1888
T in %
c
50
309929632932
1722
1640
14791453
136813231180
1034
937
3000 1500 1000n in cm-1
Fig. 2.50 (continued)
142 2 Vibrational Spectroscopy
Further Reading
1. Nakamodo K (1997) Infrared and Raman spectra of inorganic and coordination compounds.
Wiley, New York
2. Smith B (1999) Infrared spectral interpretation. CRC Press, Boca Raton
3. Long DL (2001) The Raman effect. Wiley, New York
4. Colthup ND, Daly LH, Wiberley SE (1990) Introduction to infrared and Raman spectroscopy.
Academic, New York
100
d
50
3108
T in %
3000 2000
3078
16071589 1521 1382
1479
852
1176
1163703682
1108
n in cm-1
10001500
100
T in %
e
50
30192936
1703
852
967
1365 1244
30003500 2500 2000 1500 1000
1690 1560
n in cm-1
Fig. 2.50 (a–e) Infrared spectra and some mass spectrometric data of unknown compounds.
a: Sampling technique: KBr pellet, MS: m/z in amu (Irel in%): 51 (15), 77 (49), 105 (100), 182
(32) M•+, 183 (4.2). b: Sampling technique: capillary thin film,MS: m/z in amu (Irel in%): 53 (100),
97 (4.8), 98 (0.4) M•+. c: Sampling technique: capillary thin film,MS: m/z ¼ 114 amu [M-1]•+; C6
compound. d: Sampling technique: KBr pellet, MS: m/z in amu (Irel in%): 123 (85) M•+, 124 (6).
e: Sampling technique: KBr pellet, MS: m/z in amu (Irel in%): 119 (< 0.1) M•+, 45 (100), 55 (43),
73 (82)
Further Reading 143
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